RC circuits with DC sources A Circuit i (resistors, voltage - - PowerPoint PPT Presentation

RC circuits with DC sources

C A B v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i

RC circuits with DC sources

C A B v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i C A B i v

RTh

≡

VTh

• M. B. Patil, IIT Bombay

RC circuits with DC sources

C A B v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i C A B i v

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant .

• M. B. Patil, IIT Bombay

RC circuits with DC sources

C A B v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i C A B i v

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + v → VTh = RThC dv dt + v → dv dt + v RThC = VTh RThC .

• M. B. Patil, IIT Bombay

RC circuits with DC sources

C A B v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i C A B i v

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + v → VTh = RThC dv dt + v → dv dt + v RThC = VTh RThC . * Homogeneous solution: dv dt + 1 τ v = 0 , where τ = RTh C is the “time constant.”

• V

Coul/sec × Coul V

• M. B. Patil, IIT Bombay

RC circuits with DC sources

C A B v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i C A B i v

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + v → VTh = RThC dv dt + v → dv dt + v RThC = VTh RThC . * Homogeneous solution: dv dt + 1 τ v = 0 , where τ = RTh C is the “time constant.”

• V

Coul/sec × Coul V

• → dv

v = − dt τ → log v = − t τ + K0 → v(h) = exp [(−t/τ) + K0] = K exp(−t/τ) .

• M. B. Patil, IIT Bombay

RC circuits with DC sources

C A B v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i C A B i v

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + v → VTh = RThC dv dt + v → dv dt + v RThC = VTh RThC . * Homogeneous solution: dv dt + 1 τ v = 0 , where τ = RTh C is the “time constant.”

• V

Coul/sec × Coul V

• → dv

v = − dt τ → log v = − t τ + K0 → v(h) = exp [(−t/τ) + K0] = K exp(−t/τ) . * Particular solution is a specific function that satisfies the differential equation. We know that all time derivatives will vanish as t → ∞ , making i = 0, and we get v(p) = VTh as a particular solution (which happens to be simply a constant).

• M. B. Patil, IIT Bombay

RC circuits with DC sources

C A B v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i C A B i v

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + v → VTh = RThC dv dt + v → dv dt + v RThC = VTh RThC . * Homogeneous solution: dv dt + 1 τ v = 0 , where τ = RTh C is the “time constant.”

• V

Coul/sec × Coul V

• → dv

v = − dt τ → log v = − t τ + K0 → v(h) = exp [(−t/τ) + K0] = K exp(−t/τ) . * Particular solution is a specific function that satisfies the differential equation. We know that all time derivatives will vanish as t → ∞ , making i = 0, and we get v(p) = VTh as a particular solution (which happens to be simply a constant). * v = v(h) + v(p) = K exp(−t/τ) + VTh .

• M. B. Patil, IIT Bombay

RC circuits with DC sources

C A B v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i C A B i v

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + v → VTh = RThC dv dt + v → dv dt + v RThC = VTh RThC . * Homogeneous solution: dv dt + 1 τ v = 0 , where τ = RTh C is the “time constant.”

• V

Coul/sec × Coul V

• → dv

v = − dt τ → log v = − t τ + K0 → v(h) = exp [(−t/τ) + K0] = K exp(−t/τ) . * Particular solution is a specific function that satisfies the differential equation. We know that all time derivatives will vanish as t → ∞ , making i = 0, and we get v(p) = VTh as a particular solution (which happens to be simply a constant). * v = v(h) + v(p) = K exp(−t/τ) + VTh . * In general, v(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on v.

• M. B. Patil, IIT Bombay

RC circuits with DC sources (continued)

C C A B A B i v v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i

RTh

≡

VTh

* If all sources are DC (constant), we have v(t) = A exp(−t/τ) + B , τ = RThC .

• M. B. Patil, IIT Bombay

RC circuits with DC sources (continued)

C C A B A B i v v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i

RTh

≡

VTh

* If all sources are DC (constant), we have v(t) = A exp(−t/τ) + B , τ = RThC . * i(t) = C dv dt = C × A exp(−t/τ)

• − 1

τ

• ≡ A′ exp(−t/τ) .
• M. B. Patil, IIT Bombay

RC circuits with DC sources (continued)

C C A B A B i v v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i

RTh

≡

VTh

* If all sources are DC (constant), we have v(t) = A exp(−t/τ) + B , τ = RThC . * i(t) = C dv dt = C × A exp(−t/τ)

• − 1

τ

• ≡ A′ exp(−t/τ) .

* As t → ∞, i → 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish.

• M. B. Patil, IIT Bombay

RC circuits with DC sources (continued)

C C A B A B i v v

Circuit (resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i

RTh

≡

VTh

* If all sources are DC (constant), we have v(t) = A exp(−t/τ) + B , τ = RThC . * i(t) = C dv dt = C × A exp(−t/τ)

• − 1

τ

• ≡ A′ exp(−t/τ) .

* As t → ∞, i → 0, i.e., the capacitor behaves like an open circuit since all derivatives vanish. * Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can be expressed as x(t) = K1 exp(−t/τ) + K2 , where K1 and K2 can be obtained from suitable conditions on x(t).

• M. B. Patil, IIT Bombay

Plot of f (t) = e−t/τ

t/τ e−t/τ 1 − e−t/τ 0.0 1.0 0.0 1.0 0.3679 0.6321 2.0 0.1353 0.8647 3.0 4.9787×10−2 0.9502 4.0 1.8315×10−2 0.9817 5.0 6.7379×10−3 0.9933

• M. B. Patil, IIT Bombay

Plot of f (t) = e−t/τ

t/τ e−t/τ 1 − e−t/τ 0.0 1.0 0.0 1.0 0.3679 0.6321 2.0 0.1353 0.8647 3.0 4.9787×10−2 0.9502 4.0 1.8315×10−2 0.9817 5.0 6.7379×10−3 0.9933 * For t/τ = 5, e−t/τ ≃ 0, 1 − e−t/τ ≃ 1.

• M. B. Patil, IIT Bombay

Plot of f (t) = e−t/τ

t/τ e−t/τ 1 − e−t/τ 0.0 1.0 0.0 1.0 0.3679 0.6321 2.0 0.1353 0.8647 3.0 4.9787×10−2 0.9502 4.0 1.8315×10−2 0.9817 5.0 6.7379×10−3 0.9933 * For t/τ = 5, e−t/τ ≃ 0, 1 − e−t/τ ≃ 1. * We can say that the transient lasts for about 5 time constants.

• M. B. Patil, IIT Bombay

Plot of f (t) = e−t/τ

t/τ e−t/τ 1 − e−t/τ 0.0 1.0 0.0 1.0 0.3679 0.6321 2.0 0.1353 0.8647 3.0 4.9787×10−2 0.9502 4.0 1.8315×10−2 0.9817 5.0 6.7379×10−3 0.9933 * For t/τ = 5, e−t/τ ≃ 0, 1 − e−t/τ ≃ 1. * We can say that the transient lasts for about 5 time constants.

1 1 2 3 4 5 6 x = t/τ exp(−x) 1 − exp(−x)

• M. B. Patil, IIT Bombay

Plot of f (t) = A e−t/τ + B

Plot of f (t) = A e−t/τ + B

* At t = 0, f = A + B.

Plot of f (t) = A e−t/τ + B

* At t = 0, f = A + B. * As t → ∞, f → B.

Plot of f (t) = A e−t/τ + B

* At t = 0, f = A + B. * As t → ∞, f → B. * The graph of f (t) lies between (A + B) and B. Note: If A > 0, A + B > B. If A < 0, A + B < B.

Plot of f (t) = A e−t/τ + B

* At t = 0, f = A + B. * As t → ∞, f → B. * The graph of f (t) lies between (A + B) and B. Note: If A > 0, A + B > B. If A < 0, A + B < B. * At t = 0, df dt = A e−t/τ

• − 1

τ

• = − A

τ . If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive.

Plot of f (t) = A e−t/τ + B

* At t = 0, f = A + B. * As t → ∞, f → B. * The graph of f (t) lies between (A + B) and B. Note: If A > 0, A + B > B. If A < 0, A + B < B. * At t = 0, df dt = A e−t/τ

• − 1

τ

• = − A

τ . If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive. * As t → ∞, df dt → 0, i.e., f becomes constant (equal to B).

Plot of f (t) = A e−t/τ + B

* At t = 0, f = A + B. * As t → ∞, f → B. * The graph of f (t) lies between (A + B) and B. Note: If A > 0, A + B > B. If A < 0, A + B < B. * At t = 0, df dt = A e−t/τ

• − 1

τ

• = − A

τ . If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive. * As t → ∞, df dt → 0, i.e., f becomes constant (equal to B).

5 τ t A < 0 A + B B

Plot of f (t) = A e−t/τ + B

* At t = 0, f = A + B. * As t → ∞, f → B. * The graph of f (t) lies between (A + B) and B. Note: If A > 0, A + B > B. If A < 0, A + B < B. * At t = 0, df dt = A e−t/τ

• − 1

τ

• = − A

τ . If A > 0, the derivative (slope) at t = 0 is negative; else, it is positive. * As t → ∞, df dt → 0, i.e., f becomes constant (equal to B).

5 τ t A < 0 A + B B 5 τ t A > 0 A + B B

• M. B. Patil, IIT Bombay

RL circuits with DC sources

v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L

RL circuits with DC sources

v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L i v A B L

RTh

≡

VTh

• M. B. Patil, IIT Bombay

RL circuits with DC sources

v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L i v A B L

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant .

• M. B. Patil, IIT Bombay

RL circuits with DC sources

v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L i v A B L

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + L di dt .

• M. B. Patil, IIT Bombay

RL circuits with DC sources

v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L i v A B L

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + L di dt . * Homogeneous solution: di dt + 1 τ i = 0 , where τ = L/RTh → i(h) = K exp(−t/τ) .

• M. B. Patil, IIT Bombay

RL circuits with DC sources

v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L i v A B L

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + L di dt . * Homogeneous solution: di dt + 1 τ i = 0 , where τ = L/RTh → i(h) = K exp(−t/τ) . * Particular solution is a specific function that satisfies the differential equation. We know that all time derivatives will vanish as t → ∞ , making v = 0, and we get i(p) = VTh/RTh as a particular solution (which happens to be simply a constant).

• M. B. Patil, IIT Bombay

RL circuits with DC sources

v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L i v A B L

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + L di dt . * Homogeneous solution: di dt + 1 τ i = 0 , where τ = L/RTh → i(h) = K exp(−t/τ) . * Particular solution is a specific function that satisfies the differential equation. We know that all time derivatives will vanish as t → ∞ , making v = 0, and we get i(p) = VTh/RTh as a particular solution (which happens to be simply a constant). * i = i(h) + i(p) = K exp(−t/τ) + VTh/RTh .

• M. B. Patil, IIT Bombay

RL circuits with DC sources

v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L i v A B L

RTh

≡

VTh

* If all sources are DC (constant), VTh = constant . * KVL: VTh = RTh i + L di dt . * Homogeneous solution: di dt + 1 τ i = 0 , where τ = L/RTh → i(h) = K exp(−t/τ) . * Particular solution is a specific function that satisfies the differential equation. We know that all time derivatives will vanish as t → ∞ , making v = 0, and we get i(p) = VTh/RTh as a particular solution (which happens to be simply a constant). * i = i(h) + i(p) = K exp(−t/τ) + VTh/RTh . * In general, i(t) = A exp(−t/τ) + B , where A and B can be obtained from known conditions on i.

• M. B. Patil, IIT Bombay

RL circuits with DC sources (continued)

i v A B v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L L

RTh

≡

VTh

* If all sources are DC (constant), we have i(t) = A exp(−t/τ) + B , τ = L/RTh .

• M. B. Patil, IIT Bombay

RL circuits with DC sources (continued)

i v A B v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L L

RTh

≡

VTh

* If all sources are DC (constant), we have i(t) = A exp(−t/τ) + B , τ = L/RTh . * v(t) = L di dt = L × A exp(−t/τ)

• − 1

τ

• ≡ A′ exp(−t/τ) .
• M. B. Patil, IIT Bombay

RL circuits with DC sources (continued)

i v A B v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L L

RTh

≡

VTh

* If all sources are DC (constant), we have i(t) = A exp(−t/τ) + B , τ = L/RTh . * v(t) = L di dt = L × A exp(−t/τ)

• − 1

τ

• ≡ A′ exp(−t/τ) .

* As t → ∞, v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish.

• M. B. Patil, IIT Bombay

RL circuits with DC sources (continued)

i v A B v

Circuit

A B

(resistors, voltage sources, current sources, CCVS, CCCS, VCVS, VCCS)

i L L

RTh

≡

VTh

* If all sources are DC (constant), we have i(t) = A exp(−t/τ) + B , τ = L/RTh . * v(t) = L di dt = L × A exp(−t/τ)

• − 1

τ

• ≡ A′ exp(−t/τ) .

* As t → ∞, v → 0, i.e., the inductor behaves like a short circuit since all derivatives vanish. * Since the circuit in the black box is linear, any variable (current or voltage) in the circuit can be expressed as x(t) = K1 exp(−t/τ) + K2 , where K1 and K2 can be obtained from suitable conditions on x(t).

• M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i 0 V t 5 V Vs Vs C = 1 µF Vc R = 1 k Vc(0) = 0 V

• M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i 0 V t 5 V Vs Vs C = 1 µF Vc R = 1 k Vc(0) = 0 V

* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of this change, Vc will rise. How fast can Vc change?

• M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i 0 V t 5 V Vs Vs C = 1 µF Vc R = 1 k Vc(0) = 0 V

* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of this change, Vc will rise. How fast can Vc change? * For example, what would happen if Vc changes by 1 V in 1 µs at a constant rate of 1 V /1 µs = 106 V /s?

• M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i 0 V t 5 V Vs Vs C = 1 µF Vc R = 1 k Vc(0) = 0 V

* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of this change, Vc will rise. How fast can Vc change? * For example, what would happen if Vc changes by 1 V in 1 µs at a constant rate of 1 V /1 µs = 106 V /s? * i = C dVc dt = 1 µF × 106 V s = 1 A .

• M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i 0 V t 5 V Vs Vs C = 1 µF Vc R = 1 k Vc(0) = 0 V

* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of this change, Vc will rise. How fast can Vc change? * For example, what would happen if Vc changes by 1 V in 1 µs at a constant rate of 1 V /1 µs = 106 V /s? * i = C dVc dt = 1 µF × 106 V s = 1 A . * With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL.

• M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i 0 V t 5 V Vs Vs C = 1 µF Vc R = 1 k Vc(0) = 0 V

* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of this change, Vc will rise. How fast can Vc change? * For example, what would happen if Vc changes by 1 V in 1 µs at a constant rate of 1 V /1 µs = 106 V /s? * i = C dVc dt = 1 µF × 106 V s = 1 A . * With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL. * We conclude that Vc(0+) = Vc(0−) ⇒ A capacitor does not allow abrupt changes in Vc if there is a finite resistance in the circuit.

• M. B. Patil, IIT Bombay

RC circuits: Can Vc change “suddenly?”

i 0 V t 5 V Vs Vs C = 1 µF Vc R = 1 k Vc(0) = 0 V

* Vs changes from 0 V (at t = 0−), to 5 V (at t = 0+). As a result of this change, Vc will rise. How fast can Vc change? * For example, what would happen if Vc changes by 1 V in 1 µs at a constant rate of 1 V /1 µs = 106 V /s? * i = C dVc dt = 1 µF × 106 V s = 1 A . * With i = 1 A, the voltage drop across R would be 1000 V ! Not allowed by KVL. * We conclude that Vc(0+) = Vc(0−) ⇒ A capacitor does not allow abrupt changes in Vc if there is a finite resistance in the circuit. * Similarly, an inductor does not allow abrupt changes in iL.

• M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): v(0−) = Vs(0−) = 0 V v(0+) ≃ v(0−) = 0 V Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): v(0−) = Vs(0−) = 0 V v(0+) ≃ v(0−) = 0 V Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 , A = −V0 t = 0+: 0 = A + B , t → ∞: V0 = B .

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): v(0−) = Vs(0−) = 0 V v(0+) ≃ v(0−) = 0 V Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 , A = −V0 t = 0+: 0 = A + B , t → ∞: V0 = B . v(t) = V0 [1 − exp(−t/τ)]

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): v(0−) = Vs(0−) = 0 V v(0+) ≃ v(0−) = 0 V Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 , A = −V0 t = 0+: 0 = A + B , t → ∞: V0 = B . v(t) = V0 [1 − exp(−t/τ)]

0 V

i t C v R Vs Vs V0

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): v(0−) = Vs(0−) = 0 V v(0+) ≃ v(0−) = 0 V Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 , A = −V0 t = 0+: 0 = A + B , t → ∞: V0 = B . v(t) = V0 [1 − exp(−t/τ)]

0 V

i t C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): v(0−) = Vs(0−) = 0 V v(0+) ≃ v(0−) = 0 V Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 , A = −V0 t = 0+: 0 = A + B , t → ∞: V0 = B . v(t) = V0 [1 − exp(−t/τ)]

0 V

i t C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. v(0−) = Vs(0−) = V0 v(0+) ≃ v(0−) = V0 As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): v(0−) = Vs(0−) = 0 V v(0+) ≃ v(0−) = 0 V Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 , A = −V0 t = 0+: 0 = A + B , t → ∞: V0 = B . v(t) = V0 [1 − exp(−t/τ)]

0 V

i t C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. v(0−) = Vs(0−) = V0 v(0+) ≃ v(0−) = V0 As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A + B , i.e., A = V0 , B = 0 t → ∞: 0 = B .

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): v(0−) = Vs(0−) = 0 V v(0+) ≃ v(0−) = 0 V Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. As t → ∞ , i → 0 → v(∞) = Vs(∞) = V0

Imposing (1) and (2) on Eq. (A), we get

i.e., B = V0 , A = −V0 t = 0+: 0 = A + B , t → ∞: V0 = B . v(t) = V0 [1 − exp(−t/τ)]

0 V

i t C v R Vs Vs V0

(A)

Let v(t) = A exp(−t/τ) + B, t > 0

(1) (2)

Conditions on v(t): Note that we need the condition at 0+ (and not at 0−) because Eq. (A) applies only for t > 0. v(0−) = Vs(0−) = V0 v(0+) ≃ v(0−) = V0 As t → ∞ , i → 0 → v(∞) = Vs(∞) = 0 V

Imposing (1) and (2) on Eq. (A), we get

t = 0+: V0 = A + B , i.e., A = V0 , B = 0 t → ∞: 0 = B . v(t) = V0 exp(−t/τ)

• M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0 Compute i(t), t > 0 .

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0 Compute i(t), t > 0 . (A) i(t) = C d dt V0 [1 − exp(−t/τ)] = CV0 τ exp(−t/τ) = V0 R exp(−t/τ)

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0 Compute i(t), t > 0 . (A) i(t) = C d dt V0 [1 − exp(−t/τ)] = CV0 τ exp(−t/τ) = V0 R exp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 . t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R . t → ∞: i(t) = 0 . A′ = V0 R , B′ = 0 ⇒ i(t) = V0 R exp(−t/τ)

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0 Compute i(t), t > 0 . (A) i(t) = C d dt V0 [1 − exp(−t/τ)] = CV0 τ exp(−t/τ) = V0 R exp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 . t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R . t → ∞: i(t) = 0 . A′ = V0 R , B′ = 0 ⇒ i(t) = V0 R exp(−t/τ)

0 V

i t C v R Vs Vs V0 Compute i(t), t > 0 .

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0 Compute i(t), t > 0 . (A) i(t) = C d dt V0 [1 − exp(−t/τ)] = CV0 τ exp(−t/τ) = V0 R exp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 . t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R . t → ∞: i(t) = 0 . A′ = V0 R , B′ = 0 ⇒ i(t) = V0 R exp(−t/τ)

0 V

i t C v R Vs Vs V0 Compute i(t), t > 0 . (A) i(t) = C d dt V0 [exp(−t/τ)] = −CV0 τ exp(−t/τ) = −V0 R exp(−t/τ)

RC circuits: charging and discharging transients

t

0 V

i C v R Vs Vs V0 Compute i(t), t > 0 . (A) i(t) = C d dt V0 [1 − exp(−t/τ)] = CV0 τ exp(−t/τ) = V0 R exp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 . t = 0+: v = 0 , Vs = V0 ⇒ i(0+) = V0/R . t → ∞: i(t) = 0 . A′ = V0 R , B′ = 0 ⇒ i(t) = V0 R exp(−t/τ)

0 V

i t C v R Vs Vs V0 Compute i(t), t > 0 . (A) i(t) = C d dt V0 [exp(−t/τ)] = −CV0 τ exp(−t/τ) = −V0 R exp(−t/τ)

Using these conditions, we obtain

(B) Let i(t) = A′ exp(−t/τ) + B′ , t > 0 . t → ∞: i(t) = 0 . t = 0+: v = V0 , Vs = 0 ⇒ i(0+) = −V0/R . A′ = −V0 R , B′ = 0 ⇒ i(t) = −V0 R exp(−t/τ)

• M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

t 0 V i C = 1 µF v R = 1 k Vs Vs 5 V v(t) = V0 [1 − exp(−t/τ)] i(t) = V0 R exp(−t/τ)

RC circuits: charging and discharging transients

t 0 V i C = 1 µF v R = 1 k Vs Vs 5 V v(t) = V0 [1 − exp(−t/τ)] i(t) = V0 R exp(−t/τ) 5 v (Volts) v Vs

RC circuits: charging and discharging transients

t 0 V i C = 1 µF v R = 1 k Vs Vs 5 V v(t) = V0 [1 − exp(−t/τ)] i(t) = V0 R exp(−t/τ) 5 v (Volts) v Vs 5 i (mA) time (msec) −2 2 4 6 8

RC circuits: charging and discharging transients

t 0 V i C = 1 µF v R = 1 k Vs Vs 5 V v(t) = V0 [1 − exp(−t/τ)] i(t) = V0 R exp(−t/τ) 5 v (Volts) v Vs 5 i (mA) time (msec) −2 2 4 6 8 t 0 V i C 1 µF v R = 1 k Vs Vs 5 V v(t) = V0 exp(−t/τ) i(t) = −V0 R exp(−t/τ)

RC circuits: charging and discharging transients

t 0 V i C = 1 µF v R = 1 k Vs Vs 5 V v(t) = V0 [1 − exp(−t/τ)] i(t) = V0 R exp(−t/τ) 5 v (Volts) v Vs 5 i (mA) time (msec) −2 2 4 6 8 t 0 V i C 1 µF v R = 1 k Vs Vs 5 V v(t) = V0 exp(−t/τ) i(t) = −V0 R exp(−t/τ) v (Volts) 5 v Vs

RC circuits: charging and discharging transients

t 0 V i C = 1 µF v R = 1 k Vs Vs 5 V v(t) = V0 [1 − exp(−t/τ)] i(t) = V0 R exp(−t/τ) 5 v (Volts) v Vs 5 i (mA) time (msec) −2 2 4 6 8 t 0 V i C 1 µF v R = 1 k Vs Vs 5 V v(t) = V0 exp(−t/τ) i(t) = −V0 R exp(−t/τ) v (Volts) 5 v Vs i (mA) −5 time (msec) −2 2 4 6 8

• M. B. Patil, IIT Bombay

RC circuits: charging and discharging transients

0 V 0 V 5 v (Volts) v (Volts) 5 i i t t time (msec) time (msec) −1 1 2 3 4 5 6 −1 1 2 3 4 5 6 C = 1 µF C = 1 µF v v R R Vs Vs Vs Vs 5 V 5 V R = 1 kΩ R = 100 Ω R = 1 kΩ R = 100 Ω v(t) = V0 exp(−t/τ) v(t) = V0 [1 − exp(−t/τ)]

• M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant. For example, (1) t < t1 (3) t > t2 (2) t1 < t < t2 t1 t2 Vs t

• M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant. For example, (1) t < t1 (3) t > t2 (2) t1 < t < t2 t1 t2 Vs t * For any current or voltage x(t), write general expressions such as, x(t) = A1 exp(−t/τ) + B1 , t < t1 , x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 , x(t) = A3 exp(−t/τ) + B3 , t > t2 .

• M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant. For example, (1) t < t1 (3) t > t2 (2) t1 < t < t2 t1 t2 Vs t * For any current or voltage x(t), write general expressions such as, x(t) = A1 exp(−t/τ) + B1 , t < t1 , x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 , x(t) = A3 exp(−t/τ) + B3 , t > t2 . * Work out suitable conditions on x(t) at specific time points using

• M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant. For example, (1) t < t1 (3) t > t2 (2) t1 < t < t2 t1 t2 Vs t * For any current or voltage x(t), write general expressions such as, x(t) = A1 exp(−t/τ) + B1 , t < t1 , x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 , x(t) = A3 exp(−t/τ) + B3 , t > t2 . * Work out suitable conditions on x(t) at specific time points using (a) If the source voltage/current has not changed for a “long” time (long compared to τ), all derivatives are zero. ⇒ iC = C dVc dt = 0 , and VL = L diL dt = 0 .

• M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant. For example, (1) t < t1 (3) t > t2 (2) t1 < t < t2 t1 t2 Vs t * For any current or voltage x(t), write general expressions such as, x(t) = A1 exp(−t/τ) + B1 , t < t1 , x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 , x(t) = A3 exp(−t/τ) + B3 , t > t2 . * Work out suitable conditions on x(t) at specific time points using (a) If the source voltage/current has not changed for a “long” time (long compared to τ), all derivatives are zero. ⇒ iC = C dVc dt = 0 , and VL = L diL dt = 0 . (b) When a source voltage (or current) changes, say, at t = t0 , Vc(t) or iL(t) cannot change abruptly, i.e., Vc(t+

0 ) = Vc(t− 0 ) , and iL(t+ 0 ) = iL(t− 0 ) .

• M. B. Patil, IIT Bombay

Analysis of RC/RL circuits with a piece-wise constant source

* Identify intervals in which the source voltages/currents are constant. For example, (1) t < t1 (3) t > t2 (2) t1 < t < t2 t1 t2 Vs t * For any current or voltage x(t), write general expressions such as, x(t) = A1 exp(−t/τ) + B1 , t < t1 , x(t) = A2 exp(−t/τ) + B2 , t1 < t < t2 , x(t) = A3 exp(−t/τ) + B3 , t > t2 . * Work out suitable conditions on x(t) at specific time points using (a) If the source voltage/current has not changed for a “long” time (long compared to τ), all derivatives are zero. ⇒ iC = C dVc dt = 0 , and VL = L diL dt = 0 . (b) When a source voltage (or current) changes, say, at t = t0 , Vc(t) or iL(t) cannot change abruptly, i.e., Vc(t+

0 ) = Vc(t− 0 ) , and iL(t+ 0 ) = iL(t− 0 ) .

* Compute A1, B1, · · · using the conditions on x(t).

• M. B. Patil, IIT Bombay

RL circuit: example

i v t 10 V

t0 t1 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s Find i(t).

RL circuit: example

i v t 10 V

t0 t1 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s Find i(t). (1) t < t0 (2) t0 < t < t1 (3) t > t1 There are three intervals of constant Vs:

RL circuit: example

i v t 10 V

t0 t1 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s Find i(t). (1) t < t0 (2) t0 < t < t1 (3) t > t1 There are three intervals of constant Vs: R2 R1 Vs RTh seen by L is the same in all intervals: τ = L/RTh = 0.1 s = 0.8 H/8 Ω RTh = R1 R2 = 8 Ω

RL circuit: example

i v t 10 V

t0 t1 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s Find i(t). (1) t < t0 (2) t0 < t < t1 (3) t > t1 There are three intervals of constant Vs: R2 R1 Vs RTh seen by L is the same in all intervals: τ = L/RTh = 0.1 s = 0.8 H/8 Ω RTh = R1 R2 = 8 Ω ⇒ i(t−

0 ) = 0 A ⇒ i(t+ 0 ) = 0 A .

At t = t−

0 , v = 0 V, Vs = 0 V .

RL circuit: example

i v t 10 V

t0 t1 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s Find i(t). (1) t < t0 (2) t0 < t < t1 (3) t > t1 There are three intervals of constant Vs: R2 R1 Vs RTh seen by L is the same in all intervals: τ = L/RTh = 0.1 s = 0.8 H/8 Ω RTh = R1 R2 = 8 Ω ⇒ i(t−

0 ) = 0 A ⇒ i(t+ 0 ) = 0 A .

At t = t−

0 , v = 0 V, Vs = 0 V .

10 V t v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A . If Vs did not change at t = t1, we would have t1 t0 Vs

RL circuit: example

i v t 10 V

t0 t1 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s Find i(t). (1) t < t0 (2) t0 < t < t1 (3) t > t1 There are three intervals of constant Vs: R2 R1 Vs RTh seen by L is the same in all intervals: τ = L/RTh = 0.1 s = 0.8 H/8 Ω RTh = R1 R2 = 8 Ω ⇒ i(t−

0 ) = 0 A ⇒ i(t+ 0 ) = 0 A .

At t = t−

0 , v = 0 V, Vs = 0 V .

10 V t v(∞) = 0 V, i(∞) = 10 V/10 Ω = 1 A . If Vs did not change at t = t1, we would have t1 t0 Vs i(t), t > 0 (See next slide). Using i(t+

0 ) and i(∞), we can obtain

• M. B. Patil, IIT Bombay

RL circuit: example

i v t

10 V 0.2 0.4 0.6 0.8 i (Amp) 1 time (sec)

t1 t0 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s

RL circuit: example

i v t

10 V 0.2 0.4 0.6 0.8 i (Amp) 1 time (sec)

t1 t0 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s and we need to work out the solution for t > t1 separately. In reality, Vs changes at t = t1,

RL circuit: example

i v t

10 V 0.2 0.4 0.6 0.8 i (Amp) 1 time (sec)

t1 t0 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s and we need to work out the solution for t > t1 separately. In reality, Vs changes at t = t1, Consider t > t1. For t0 < t < t1, i(t) = 1 − exp(−t/τ) Amp.

RL circuit: example

i v t

10 V 0.2 0.4 0.6 0.8 i (Amp) 1 time (sec)

t1 t0 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s and we need to work out the solution for t > t1 separately. In reality, Vs changes at t = t1, Consider t > t1. For t0 < t < t1, i(t) = 1 − exp(−t/τ) Amp. i(t+

1 ) = i(t− 1 ) = 1 − e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A.

RL circuit: example

i v t

10 V 0.2 0.4 0.6 0.8 i (Amp) 1 time (sec)

t1 t0 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s and we need to work out the solution for t > t1 separately. In reality, Vs changes at t = t1, Consider t > t1. For t0 < t < t1, i(t) = 1 − exp(−t/τ) Amp. i(t+

1 ) = i(t− 1 ) = 1 − e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A. Let i(t) = A exp(−t/τ) + B.

RL circuit: example

i v t

10 V 0.2 0.4 0.6 0.8 i (Amp) 1 time (sec)

t1 t0 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s and we need to work out the solution for t > t1 separately. In reality, Vs changes at t = t1, Consider t > t1. For t0 < t < t1, i(t) = 1 − exp(−t/τ) Amp. i(t+

1 ) = i(t− 1 ) = 1 − e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A. Let i(t) = A exp(−t/τ) + B. It is convenient to rewrite i(t) as i(t) = A′ exp[−(t − t1)/τ] + B.

RL circuit: example

i v t

10 V 0.2 0.4 0.6 0.8 i (Amp) 1 time (sec)

t1 t0 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s and we need to work out the solution for t > t1 separately. In reality, Vs changes at t = t1, Consider t > t1. For t0 < t < t1, i(t) = 1 − exp(−t/τ) Amp. i(t+

1 ) = i(t− 1 ) = 1 − e−1 = 0.632 A (Note: t1/τ = 1).

i(∞) = 0 A. Let i(t) = A exp(−t/τ) + B. It is convenient to rewrite i(t) as i(t) = A′ exp[−(t − t1)/τ] + B. Using i(t+

1 ) and i(∞), we get

i(t) = 0.632 exp[−(t − t1)/τ] A.

• M. B. Patil, IIT Bombay

RL circuit: example

i v

t

10 V 0.2 0.4 0.6 0.8 i (Amp) 1 time (sec)

i(t) = 0.632 exp[−(t − t1)/τ] A. t1 t0 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s

RL circuit: example

i v

t

10 V 0.2 0.4 0.6 0.8 i (Amp) 1 time (sec)

i(t) = 0.632 exp[−(t − t1)/τ] A. t1 t0 R2 R1 Vs Vs R1 = 10 Ω R2 = 40 Ω L = 0.8 H t0 = 0 t1 = 0.1 s

(SEQUEL file: ee101_rl1.sqproj) 0.2 0.4 0.6 0.8 1 i (Amp) time (sec)

Combining the solutions for t0 < t < t1 and t > t1, we get

• M. B. Patil, IIT Bombay

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0 i 5 k 1 k 6 V

R2 R1 ic R3 = 5 k vc 5 µF

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0 i 5 k 1 k 6 V

R2 R1 ic R3 = 5 k vc 5 µF

AND i i 5 k 1 k 5 k 5 k 5 k 6 V

ic t < 0 ic vc t > 0 vc 5 µF 5 µF

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0 i 5 k 1 k 6 V

R2 R1 ic R3 = 5 k vc 5 µF

AND i i 5 k 1 k 5 k 5 k 5 k 6 V

ic t < 0 ic vc t > 0 vc 5 µF 5 µF t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA.

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0 i 5 k 1 k 6 V

R2 R1 ic R3 = 5 k vc 5 µF

AND i i 5 k 1 k 5 k 5 k 5 k 6 V

ic t < 0 ic vc t > 0 vc 5 µF 5 µF t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA. vc(0−) = i(0−) R1 = 5 V ⇒ vc(0+) = 5 V

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0 i 5 k 1 k 6 V

R2 R1 ic R3 = 5 k vc 5 µF

AND i i 5 k 1 k 5 k 5 k 5 k 6 V

ic t < 0 ic vc t > 0 vc 5 µF 5 µF t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA. vc(0−) = i(0−) R1 = 5 V ⇒ vc(0+) = 5 V ⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA.

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0 i 5 k 1 k 6 V

R2 R1 ic R3 = 5 k vc 5 µF

AND i i 5 k 1 k 5 k 5 k 5 k 6 V

ic t < 0 ic vc t > 0 vc 5 µF 5 µF t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA. vc(0−) = i(0−) R1 = 5 V ⇒ vc(0+) = 5 V ⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA. Let i(t) = A exp(-t/τ) + B for t > 0, with τ = 10 k × 5 µF = 50 ms.

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0 i 5 k 1 k 6 V

R2 R1 ic R3 = 5 k vc 5 µF

AND i i 5 k 1 k 5 k 5 k 5 k 6 V

ic t < 0 ic vc t > 0 vc 5 µF 5 µF t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA. vc(0−) = i(0−) R1 = 5 V ⇒ vc(0+) = 5 V ⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA. Let i(t) = A exp(-t/τ) + B for t > 0, with τ = 10 k × 5 µF = 50 ms. i(t) = 0.5 exp(-t/τ) mA. Using i(0+) and i(∞) = 0 A, we get

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0 i 5 k 1 k 6 V

R2 R1 ic R3 = 5 k vc 5 µF

AND i i 5 k 1 k 5 k 5 k 5 k 6 V

ic t < 0 ic vc t > 0 vc 5 µF 5 µF t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA. vc(0−) = i(0−) R1 = 5 V ⇒ vc(0+) = 5 V ⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA. Let i(t) = A exp(-t/τ) + B for t > 0, with τ = 10 k × 5 µF = 50 ms. i(t) = 0.5 exp(-t/τ) mA. Using i(0+) and i(∞) = 0 A, we get

1 time (sec) i (mA) 0.5

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0 i 5 k 1 k 6 V

R2 R1 ic R3 = 5 k vc 5 µF

AND i i 5 k 1 k 5 k 5 k 5 k 6 V

ic t < 0 ic vc t > 0 vc 5 µF 5 µF t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA. vc(0−) = i(0−) R1 = 5 V ⇒ vc(0+) = 5 V ⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA. Let i(t) = A exp(-t/τ) + B for t > 0, with τ = 10 k × 5 µF = 50 ms. i(t) = 0.5 exp(-t/τ) mA. Using i(0+) and i(∞) = 0 A, we get

1 time (sec) i (mA) 0.5 −0.5 5 time (sec) time (sec) 0.5 0.5

ic (mA) vc (V)

RC circuit: The switch has been closed for a long time and opens at t = 0.

t=0 i 5 k 1 k 6 V

R2 R1 ic R3 = 5 k vc 5 µF

AND i i 5 k 1 k 5 k 5 k 5 k 6 V

ic t < 0 ic vc t > 0 vc 5 µF 5 µF t = 0−: capacitor is an open circuit ⇒ i(0−) = 6 V/(5 k + 1 k) = 1 mA. vc(0−) = i(0−) R1 = 5 V ⇒ vc(0+) = 5 V ⇒ i(0+) = 5 V/(5 k + 5 k) = 0.5 mA. Let i(t) = A exp(-t/τ) + B for t > 0, with τ = 10 k × 5 µF = 50 ms. i(t) = 0.5 exp(-t/τ) mA. Using i(0+) and i(∞) = 0 A, we get

1 time (sec) i (mA) 0.5 −0.5 5 time (sec) time (sec) 0.5 0.5

ic (mA) vc (V)

(SEQUEL file: ee101_rc2.sqproj)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R iC (mA) VS, VC (Volts) Vc 5 k iC C 1 µF 2 1 −1 10 20 10 30 40 50 t (msec)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R iC (mA) VS, VC (Volts) Vc 5 k iC C 1 µF 2 1 −1 10 20 10 30 40 50 t (msec) Find expressions for VC (t) and iC (t) in steady state (in terms of R, C, V0, T1, T2). V0 T1 T2

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0.

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

V (2)

C (T1) = V2, V (2) C (∞) = 0

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

V (2)

C (T1) = V2, V (2) C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ.

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

V (2)

C (T1) = V2, V (2) C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ. V (2)

C (t) = V2e−(t−T1)/τ

(2)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

V (2)

C (T1) = V2, V (2) C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ. V (2)

C (t) = V2e−(t−T1)/τ

(2) Now use the conditions: V (1)

C (T1) = V2, V (2) C (T1 + T2) = V1.

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

V (2)

C (T1) = V2, V (2) C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ. V (2)

C (t) = V2e−(t−T1)/τ

(2) Now use the conditions: V (1)

C (T1) = V2, V (2) C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

V (2)

C (T1) = V2, V (2) C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ. V (2)

C (t) = V2e−(t−T1)/τ

(2) Now use the conditions: V (1)

C (T1) = V2, V (2) C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3) V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

V (2)

C (T1) = V2, V (2) C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ. V (2)

C (t) = V2e−(t−T1)/τ

(2) Now use the conditions: V (1)

C (T1) = V2, V (2) C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3) V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4) Rewrite with a ≡ e−T1/τ, b ≡ e−T2/τ.

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

V (2)

C (T1) = V2, V (2) C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ. V (2)

C (t) = V2e−(t−T1)/τ

(2) Now use the conditions: V (1)

C (T1) = V2, V (2) C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3) V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4) Rewrite with a ≡ e−T1/τ, b ≡ e−T2/τ. V2 = −(V0 − V1)a + V0 (5)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

V (2)

C (T1) = V2, V (2) C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ. V (2)

C (t) = V2e−(t−T1)/τ

(2) Now use the conditions: V (1)

C (T1) = V2, V (2) C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3) V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4) Rewrite with a ≡ e−T1/τ, b ≡ e−T2/τ. V2 = −(V0 − V1)a + V0 (5) V1 = b V2 (6)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 0 < t < T1 Let V (1)

C (t) = A e−t/τ + B

V (1)

C (0) = V1, V (1) C (∞) = V0

→ B = V0, A = V1 − V0. V (1)

C (t) = −(V0 − V1)e−t/τ + V0

(1) T1 < t < T2 Let V (2)

C (t) = A′ e−t/τ + B′

V (2)

C (T1) = V2, V (2) C (∞) = 0

→ B′ = 0, A′ = V2 eT1/τ. V (2)

C (t) = V2e−(t−T1)/τ

(2) Now use the conditions: V (1)

C (T1) = V2, V (2) C (T1 + T2) = V1.

V2 = −(V0 − V1)e−T1/τ + V0 (3) V1 = V2e−(T1+T2−T1)/τ = V2e−T2/τ (4) Rewrite with a ≡ e−T1/τ, b ≡ e−T2/τ. V2 = −(V0 − V1)a + V0 (5) V1 = b V2 (6) Solve to get V1 = b V0 1 − a 1 − ab , V2 = V0 1 − a 1 − ab

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 V1 = b V0 1 − a 1 − ab , V2 = V0 1 − a 1 − ab , with a = e−T1/τ, b = e−T2/τ.

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 V1 = b V0 1 − a 1 − ab , V2 = V0 1 − a 1 − ab , with a = e−T1/τ, b = e−T2/τ. V (1)

C (t) = −(V0 − V1)e−t/τ + V0,

V (2)

C (t) = V2e−(t−T1)/τ.

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 V1 = b V0 1 − a 1 − ab , V2 = V0 1 − a 1 − ab , with a = e−T1/τ, b = e−T2/τ. V (1)

C (t) = −(V0 − V1)e−t/τ + V0,

V (2)

C (t) = V2e−(t−T1)/τ.

Current calculation:

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 V1 = b V0 1 − a 1 − ab , V2 = V0 1 − a 1 − ab , with a = e−T1/τ, b = e−T2/τ. V (1)

C (t) = −(V0 − V1)e−t/τ + V0,

V (2)

C (t) = V2e−(t−T1)/τ.

Current calculation: Method 1: iC (t) = C dVC dt (home work)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 V1 = b V0 1 − a 1 − ab , V2 = V0 1 − a 1 − ab , with a = e−T1/τ, b = e−T2/τ. V (1)

C (t) = −(V0 − V1)e−t/τ + V0,

V (2)

C (t) = V2e−(t−T1)/τ.

Current calculation: Method 1: iC (t) = C dVC dt (home work) Method 2: Start from scratch!

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2 V1 = b V0 1 − a 1 − ab , V2 = V0 1 − a 1 − ab , with a = e−T1/τ, b = e−T2/τ. V (1)

C (t) = −(V0 − V1)e−t/τ + V0,

V (2)

C (t) = V2e−(t−T1)/τ.

Current calculation: Method 1: iC (t) = C dVC dt (home work) Method 2: Start from scratch!

STOP

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1.

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

i(2)

C (T1) = −I2, i(2) C (∞) = 0

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

i(2)

C (T1) = −I2, i(2) C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ.

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

i(2)

C (T1) = −I2, i(2) C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ. i(2)

C (t) = −I2e−(t−T1)/τ

(2)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

i(2)

C (T1) = −I2, i(2) C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ. i(2)

C (t) = −I2e−(t−T1)/τ

(2) Now use the conditions: i(1)

C (T1) − i(2) C (T1) = ∆ = V0/R,

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

i(2)

C (T1) = −I2, i(2) C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ. i(2)

C (t) = −I2e−(t−T1)/τ

(2) Now use the conditions: i(1)

C (T1) − i(2) C (T1) = ∆ = V0/R,

i(1)

C (0) − i(2) C (T1 + T2) = ∆ = V0/R.

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

i(2)

C (T1) = −I2, i(2) C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ. i(2)

C (t) = −I2e−(t−T1)/τ

(2) Now use the conditions: i(1)

C (T1) − i(2) C (T1) = ∆ = V0/R,

i(1)

C (0) − i(2) C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

i(2)

C (T1) = −I2, i(2) C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ. i(2)

C (t) = −I2e−(t−T1)/τ

(2) Now use the conditions: i(1)

C (T1) − i(2) C (T1) = ∆ = V0/R,

i(1)

C (0) − i(2) C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3) I1 − (−I2e−(T1+T2−T1)/τ) = ∆ (4)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

i(2)

C (T1) = −I2, i(2) C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ. i(2)

C (t) = −I2e−(t−T1)/τ

(2) Now use the conditions: i(1)

C (T1) − i(2) C (T1) = ∆ = V0/R,

i(1)

C (0) − i(2) C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3) I1 − (−I2e−(T1+T2−T1)/τ) = ∆ (4) a I1 + I2 = ∆ (5)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

i(2)

C (T1) = −I2, i(2) C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ. i(2)

C (t) = −I2e−(t−T1)/τ

(2) Now use the conditions: i(1)

C (T1) − i(2) C (T1) = ∆ = V0/R,

i(1)

C (0) − i(2) C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3) I1 − (−I2e−(T1+T2−T1)/τ) = ∆ (4) a I1 + I2 = ∆ (5) I1 + b I2 = ∆ (6)

• M. B. Patil, IIT Bombay

RC circuit: example

VS R C Vc iC I1 −I2 iC ∆ VC VS V0 V2 V1 T1 T2 T1 T2 ∆ 0 < t < T1 Let i(1)

C (t) = A e−t/τ + B

i(1)

C (0) = I1, i(1) C (∞) = 0

→ B = 0, A = I1. i(1)

C (t) = I1e−t/τ

(1) T1 < t < T2 Let i(2)

C (t) = A′ e−t/τ + B′

i(2)

C (T1) = −I2, i(2) C (∞) = 0

→ B′ = 0, A′ = −I2 eT1/τ. i(2)

C (t) = −I2e−(t−T1)/τ

(2) Now use the conditions: i(1)

C (T1) − i(2) C (T1) = ∆ = V0/R,

i(1)

C (0) − i(2) C (T1 + T2) = ∆ = V0/R.

I1e−T1/τ − (−I2) = ∆ (3) I1 − (−I2e−(T1+T2−T1)/τ) = ∆ (4) a I1 + I2 = ∆ (5) I1 + b I2 = ∆ (6) Solve to get I1 = ∆ 1 − b 1 − ab , I2 = ∆ 1 − a 1 − ab (a = e−T1/τ, b = e−T2/τ)

• M. B. Patil, IIT Bombay

RC circuit: example

• Q

Q insulator conductor conductor Unit: Farad (F)

VS R VC C Vc iC C = ǫ A t t iC iC iC = dQ dt = C dVC dt I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2

Charge conservation:

• M. B. Patil, IIT Bombay

RC circuit: example

• Q

Q insulator conductor conductor Unit: Farad (F)

VS R VC C Vc iC C = ǫ A t t iC iC iC = dQ dt = C dVC dt I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2

Charge conservation: Periodic steady state: All quantities are periodic, i.e., x(t0 + T) = x(t0)

• M. B. Patil, IIT Bombay

RC circuit: example

• Q

Q insulator conductor conductor Unit: Farad (F)

VS R VC C Vc iC C = ǫ A t t iC iC iC = dQ dt = C dVC dt I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2

Charge conservation: Periodic steady state: All quantities are periodic, i.e., x(t0 + T) = x(t0) Capacitor charge: Q(t0 + T) = Q(t0)

• M. B. Patil, IIT Bombay

RC circuit: example

• Q

Q insulator conductor conductor Unit: Farad (F)

VS R VC C Vc iC C = ǫ A t t iC iC iC = dQ dt = C dVC dt I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2

Charge conservation: Periodic steady state: All quantities are periodic, i.e., x(t0 + T) = x(t0) Capacitor charge: Q(t0 + T) = Q(t0) iC = dQ dt → Q =

• iC dt.
• M. B. Patil, IIT Bombay

RC circuit: example

• Q

Q insulator conductor conductor Unit: Farad (F)

VS R VC C Vc iC C = ǫ A t t iC iC iC = dQ dt = C dVC dt I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2

Charge conservation: Periodic steady state: All quantities are periodic, i.e., x(t0 + T) = x(t0) Capacitor charge: Q(t0 + T) = Q(t0) iC = dQ dt → Q =

• iC dt.

Q(t0 + T) = Q(t0) → Q(t0 + T) − Q(t0) = 0 → t0+T

t0

iC dt = 0.

• M. B. Patil, IIT Bombay

RC circuit: example

• Q

Q insulator conductor conductor Unit: Farad (F)

VS R VC C Vc iC C = ǫ A t t iC iC iC = dQ dt = C dVC dt I1 −I2 iC VC VS V0 V2 V1 T1 T2 T1 T2

Charge conservation: Periodic steady state: All quantities are periodic, i.e., x(t0 + T) = x(t0) Capacitor charge: Q(t0 + T) = Q(t0) iC = dQ dt → Q =

• iC dt.

Q(t0 + T) = Q(t0) → Q(t0 + T) − Q(t0) = 0 → t0+T

t0

iC dt = 0. T iC dt = 0 → T1 iC dt + T1+T2

T1

iC dt = 0 → T1+T2

T1

iC dt = − T1 iC dt.

• M. B. Patil, IIT Bombay

−10 10 2 4 6 8 10 1 2 3 4 5 6 t (msec)

VS R Vc 1 k iC C 1 µF T1 = 1.5 msec, T2 = 1.5 msec. V1 = 1.8 V, V2 = 8.2 V. I1 = 8.2 mA, I2 = 8.2 mA.

VS VC iC (mA)

−10 10 2 4 6 8 10 1 2 3 4 5 6 t (msec)

VS R Vc 1 k iC C 1 µF T1 = 1.5 msec, T2 = 1.5 msec. V1 = 1.8 V, V2 = 8.2 V. I1 = 8.2 mA, I2 = 8.2 mA.

VS VC iC (mA)

t (msec) 1 2 3 4 5 6

T1 = 1 msec, T2 = 2 msec. V1 = 0.9 V, V2 = 6.7 V. I1 = 9.1 mA, I2 = 6.7 mA.

−10 10 2 4 6 8 10 1 2 3 4 5 6 t (msec)

VS R Vc 1 k iC C 1 µF T1 = 1.5 msec, T2 = 1.5 msec. V1 = 1.8 V, V2 = 8.2 V. I1 = 8.2 mA, I2 = 8.2 mA.

VS VC iC (mA)

t (msec) 1 2 3 4 5 6

T1 = 1 msec, T2 = 2 msec. V1 = 0.9 V, V2 = 6.7 V. I1 = 9.1 mA, I2 = 6.7 mA.

t (msec) 1 2 3 4 5 6

T1 = 2 msec, T2 = 1 msec. V1 = 3.4 V, V2 = 9.1 V. I1 = 6.7 mA, I2 = 9.1 mA.

−10 10 2 4 6 8 10 1 2 3 4 5 6 t (msec)

VS R Vc 1 k iC C 1 µF T1 = 1.5 msec, T2 = 1.5 msec. V1 = 1.8 V, V2 = 8.2 V. I1 = 8.2 mA, I2 = 8.2 mA.

VS VC iC (mA)

t (msec) 1 2 3 4 5 6

T1 = 1 msec, T2 = 2 msec. V1 = 0.9 V, V2 = 6.7 V. I1 = 9.1 mA, I2 = 6.7 mA.

t (msec) 1 2 3 4 5 6

T1 = 2 msec, T2 = 1 msec. V1 = 3.4 V, V2 = 9.1 V. I1 = 6.7 mA, I2 = 9.1 mA. SEQUEL file: ee101 rc1b.sqproj

• M. B. Patil, IIT Bombay

t (msec) 1 2 3 4 5 6 2 4 6 8 10 −100 100

VS R T1 = 1 msec, T2 = 2 msec. V1 ≈ 0 V, V2 = 10 V. I1 = 100 mA, I2 = 100 mA. Vc 0.1 k iC C 1 µF

iC (mA) VS VC

• M. B. Patil, IIT Bombay