Electric circuit components Capacitor stores charge and potential - - PDF document

Electric circuit components

Capacitor – stores charge and potential energy, measured in Farads (F) Battery – generates a constant electrical potential difference (∆V) across it. Measured in Volts (V). Resistor – resists flow of charge due to scattering; dissipates energy. Measured in Ohms (Ω)

Direct Current (DC) circuits

First, we will consider circuits with batteries and resistors.

R V = 5 V I

Resistors in Series

I I I I = = =

3 2 1

3 3 2 2 1 1

IR V V IR V V IR V V

yb xy ax

= = = = = =

( )

3 2 1 3 2 1

R R R I V V V Vab + + = + + =

The current through all resistors is the same

3 2 1

R R R R IR V

eq eq ab

+ + = → =

Resistors in Series

∑

=

i i eq

R R

Adding resistors in series always increases the total resistance.

Resistors in Series: Example

Calculate I and V across each resistor

V RA RB Ibat IA IB

RA = 1.0 Ω RB = 3.0 Ω V = 6 Volts

V RA RB Ibat IA IB

Two different wires are connected to a battery in series (in a chain, one after another). How does the current in upper wire A compare to the current in lower wire B? A) iA > iB B) iA < iB C) iA = iB D) Impossible to determine.

Clicker Question

What is the equivalent resistance?

Ω = Ω + Ω = + = . 4 . 3 . 1

2 1

R R Req

What is the current through the circuit? Assuming everything is Ohmic, V=IR

A V R V i i i

eq B A

5 . 1 . 4 6 = Ω = = = =

V RA RB Ibat IA IB

Resistors in Series: Example

What about the Voltage change across each resistor?

V RA RB Ibat IA IB

RA = 1.0 Ω, iA

A=1.5 A

=1.5 A RB = 3.0 Ω, iB = 1.5 A V = 6 Volts

V A R i V V

A A A A

5 . 1 ) . 1 )( 5 . 1 ( = Ω = = ∆ = V A R i V V

B B B B

5 . 4 ) . 3 )( 5 . 1 ( = Ω = = ∆ =

Resistors in Series: Example

V RA RB Ibat IA IB

Clicker Question

If we consider the Voltage change starting at Point P and going all the way around the circuit loop back to Point P, what is the total ∆V? Α)∆V = +6.0 Volts Β)∆V = +1.5 Volts C) ∆V = -4.5 Volts D) ∆V = 0.0 Volts E) ∆V = -6.0 Volts Point P

V RA RB Ibat IA IB

P1 P2 P3 P4 P5

Position In Loop

Voltage 1 2 3 4 5 1

6V 4.5V 3.0V 1.5V 0.0V

Define V=0 at P1.

Resistors in Series: Example

Clicker Question

We start with the left circuit with one lightbulb (A). The brightness of the bulb directly reflects the power. If we add a second bulb (B) as shown on the right, what happens to the bulbs?

V A V A B

A)Bulb A is equally bright. B)Bulb A is dimmer than before C)Bulb A is brighter than before Two light bulbs, A and B, are in series, so they carry the same current. Light bulb A is brighter than B. Which bulb has higher resistance? A) A B) B C) Same resistance.

V A B

Answer: Bulb A has higher resistance. Since the resistors are in series, they have the same current I. According to P = I2 R, if I = constant, then higher R gives higher P.

Clicker Question

Clicker Question

If we connect a battery with V = 6 Volts with three resistors R1 = 10 Ω, R2 = 20 Ω, R3 = 30 Ω, which of the following is true? A)The current through each resistor is the same. B)The Voltage change through each resistor is the same. C)The resistance of each resistor is the same. D)None of the above are true.

Resistors in Parallel

V V V V V

ab =

= = =

3 2 1

3 3 2 2 1 1

/ / / R V I R V I R V I = = =

⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + + = + + =

3 2 1 3 2 1

1 1 1 R R R V I I I I

The voltage across all resistors is the same

3 2 1

1 1 1 1

/

R R R R eq

eq

R V I + + = → =

Resistors in Parallel

∑

=

i i eq

R R 1 1

Adding resistors in parallel always decreases the total resistance.

Traffic analog

Resistors in parallel are like having more lanes

• n the highway.

This reduces the resistance for getting from one place to another.

Resistors in parallel: Example

R1 R2 R3

V

i i1 i2 i3

3 2 1

I I I I + + =

If all three resistors were the same then:

I I I I 3 1

3 2 1

= = =

Resistor configurations

3 2 1

1 1 1 1 R R R Req + + =

3 2 1

R R R Req + + =

Resistors in Series Resistors in Parallel Always increases the resistance! Always decreases the resistance!

V=12V

Clicker Question

V=12V

A A B We start with the left circuit with one lightbulb (A). The brightness of the bulb directly reflects the power. If we add a second identical bulb (B) as shown on the right, what happens to the bulbs? A)Bulb A is equally bright. B)Bulb A is dimmer than before C)Bulb A is brighter than before

Clicker Question

If each of these six light bulbs is identical, which bulb is going to be the brightest? A)Bulb A B)Bulb B C)Bulb C D)Bulb D E)Bulb E

Clicker Question

The three light bulbs A, B, and C are identical. How does the brightness of bulbs B and C together compare with the brightness of bulb A?

V=12V A B C

A) Total power in B+C = power in A. B) Total power in B+C > power in A. C) Total power in B+C < power in A. Answer: Use P = V2/Rtot For bulbs B and C, Rtot = 2R. Total power in B+C < power in A.

Clicker Question

V=12V 1 2 3

In the circuit below, what happens to the brightness of bulb 1, when bulb 2 burns out? (When a bulb burns out, its resistance becomes infinite.) A) Bulb 1 gets brighter B) Bulb 1 gets dimmer. C) It's brightness remains the same. (Hint: What happens to the current from the battery when bulb 2 burns out.)

Answers: Bulb 1 gets dimmer! When bulb 2 burns outs, the filament inside breaks and R2 becomes infinitely large. The total equivalent resistance which the battery sees increases (since bulb 2 is gone, there are fewer paths for the current flow, so less flow, more total resistance.) Since the battery sees a larger R-tot, the current from the battery I-tot = V/Rtot is reduced. Less current from the battery means less current through bulb 1, less light.

V=12V 1 2 3

Which is the best way to wire a house? CPS question (A) (B)

Resistor networks Resistor networks

Kirchhoff’s rules

Kirchhoff’s Voltage Loop Law: The change in Voltage around any closed loop must be zero. Kirchhoff’s Current Junction Law: In steady state, the current going into a junction (or point) must equal the current going out of that junction (or point).

V=12V

Clicker Question

What is the electric potential difference across the upper light bulb (resistor)? Think about our Voltage Loop Rule. A)|V| = 0 Volts B)|V| = 6 Volts C)|V| = 12 Volts D)|V| = 24 Volts E)None of the above answers.

Kirchhoff’s rules: Example

Example 1: Choose R1 = R2 = R3 = 10 Ω

V=12V 1 2 3

• 1. Longer line on battery is

higher voltage and the shorter line is lower voltage, by convention. +

• 2. We label the current in

each section as shown. Note that we can guess the direction at this point. The answer will be independent

• f our guess.

ib ib ib ib i2 i3 i1

V=12V 1 2 3

+

• ib

ib ib ib i2 i3 i1

• 3. Apply the Voltage Loop

Rule around each possible loop in the circuit. Make sure to label the direction of your loop! Make sure to pick a starting point to go around the loop. Then add up the Voltage changes around the loop.

Kirchhoff’s rules: Example

Critical Sign Convention! Critical Sign Convention!

V=12V

• +

If your loop goes through a battery from – to + the Voltage increases (e.g. ∆V = +12 V)

V=12V

• +

If your loop goes through a battery from + to – the Voltage decreases (e.g. ∆V = -12 V)

Critical Sign Convention! Critical Sign Convention!

2

i If you go across a resistor and the loop direction and guessed current direction are the same, the voltage decreases (e.g. ∆V = -iR)

2

i If you go across a resistor and the loop direction and guessed current direction are

• pposite, the voltage increases (e.g. ∆V = +iR)

V=12V 1 2 3

+

• ib

ib ib ib i2 i3 i1 Starting at Point P go around the loop. P

V Vbat 12 + = ∆

3 3 3

R i VR − = ∆ 12

1 1 3 3

= − − + = ∆ R i R i Vsum

1 1 1

R i VR − = ∆

Kirchhoff’s rules: Example

V=12V 1 2 3

+

• ib

ib ib i2 i3 i1 P Now try other path starting at Point P around the new loop.

V Vbat 12 + = ∆

2 2 2

R i VR − = ∆ 12

1 1 2 2

= − − + = ∆ R i R i Vsum

1 1 1

R i VR − = ∆

Kirchhoff’s rules: Example

12

1 1 2 2

= − − + = ∆ R i R i Vsum 12

1 1 3 3

= − − + = ∆ R i R i Vsum

For our specific circuit, we used R1 = R2 = R3 = 10 Ω

10 10 12

1 2

= − − i i 10 10 12

1 3

= − − i i

We have two equations and three unknowns. We need additional information.

Kirchhoff’s rules: Example

V=12V 1 2 3

+

• ib

ib ib i3 i1 i2 Q Apply the current junction rule at point Q below. 1 3 2

i i i i i

• ut

in

= + =

ib Also, (as we could have done at the start): b

i i =

1

Kirchhoff’s rules: Example

10 10 12

1 2

= − − i i 10 10 12

1 3

= − − i i

1 3 2

i i i = +

Solve 3 equations and 3 unknowns.

A i A i A i 4 . 4 . 8 .

3 2 1

= = =

Kirchhoff’s rules: Example

V=12V 1 2 3

+

• i1

i1 i1 i1 i2 i3 i1

A i A i A i 4 . 4 . 8 .

3 2 1

= = =

Does this make sense? i2 = i3 since they are two parallel, equal resistance paths. Current from R2 and R3 must go into R1.

Kirchhoff’s rules: Example

Clicker Question

V=12V 1 2 3

+

• i1

i1 i1 i1 i2 i3 i1 Amps i Amps i Amps i 4 . 4 . 8 .

3 2 1

= = = Ω = Ω = Ω = 10 10 10

3 2 1

R R R What is the value of the equivalent resistor that would replace the three resistors below? A)Req = 5 Ω B)Req = 10 Ω C)Req = 15 Ω D)Req = 30 Ω E)None of the Above

V=12V 1 2 3

+

• i1

i1 i1 i1 i2 i3 i1

Ω = Ω = Ω = 10 10 10

3 2 1

R R R

Two ways to think about it.

• 1. Battery of 12 V has 0.8 Amps

coming out of it. By V=iR the Req = 12V/0.8A=15 Ω

• 2. R2 and R3 are in parallel.

Then R23 and R1 are in series.

Amps i Amps i Amps i 4 . 4 . 8 .

3 2 1

= = =

Ω = ⎟ ⎟ ⎠ ⎞ ⎜ ⎜ ⎝ ⎛ + = 5 1 1 1

3 2 23

R R R

Ω = + = = 15

23 1 123

R R R Req

Which of these diagrams represent the same circuit?

Clicker Question

• A. a and b
• B. a and c
• C. b and c
• D. a, b, and c
• E. a, b, and d

A circuit with two batteries is shown below. The directions of the currents have been chosen (guessed) as shown. Which is the correct current equation for this circuit? A)I2 = I1 + I3 B) I1 = I2 + I3 C) I3 = I1 + I2 D) None of these.

R2 R1 R3 V1 V2

2

I1 I3 I Loop 1.

Clicker Question

Clicker Question

R2 R1 R3 V1 V2

2

I1 I3 I Loop 1. I3

Which equation below is the correct equation for Loop 1? A) –V2 + I1R1 – I2R2 = 0 B) V2 + I1R1 – I2R2 = 0 C) –V2 – I1R1 + I2R2 = 0 D) V2 + I1R1 + I2R2 = 0 E) None of these. Answer: –V2 + I1R1 – I2R2 = 0

R2 R1 R3 V1 V2

2

I1 I3 I Loop 1. I3

Finish a full analysis of this circuit. Consider case: V1 = 12 V, V2 = 24 V, R1=R2=R3 = 1 Ω.

2 1 3

i i i + =

2 2 1 1 2

= − + − R i R i V 24

2 1

= − + − i i

3 3 2 2 1

= − − + R i R i V 12

3 2

= − − + i i

Loop 2

A i A i A i 16 , 4 , 20

3 2 1

= − = =

Two light bulbs A and B are connected in series to a constant voltage source. When a wire is connected across B as shown, the brightness of bulb A ... A: increases B: decreases, but remains glowing C: decreases to zero (bulb A goes completely dark, no current) D: remains unchanged Answer: Bulb A increases in brightness (since the current increases when B is shorted).

Clicker Question Circuit Probes

An Ammeter measures the current through itself. An ideal Ammeter has zero resistance so it does not affect the circuit! A i i To measure the current through R, must place Ammeter in series.

R V = 5 V I

A

Consider the circuit shown. If you want to measure the current thru bulb 3, how should the ammeter be attached (assume you

• nly attach one at a time)?

V 4 1 2 3 A A A A Pink Yellow Green Blue

A) Pink B) Yellow C) Green D) Blue E) Two of the 4 positions can be used to measure bulb3’s current. Answer: B) Yellow

Clicker Question Circuit Probes

To measure voltage across R, must place Voltmeter in parallel with R. You want the Voltmeter to be very high resistance (ideally infinity) to avoid drawing current.

R V = 5 V I

V

In the circuit shown, what does the voltmeter read?

6 V 2Ω V

A: 6 V B: 3 V C: 2 V D: 0 V E: Voltmeter will "fry"

Clicker Question Clicker Question

6 V 2Ω A Now you switch the voltmeter over to "amp" mode. (But you leave it in the same position in the circuit) What does the Ammeter read? A: 6 A B: 3 A C: 2 A D: 0 A E: Ammeter will "fry"

Ammeter will fry. Ideal Ammeter has zero internal resistance. If you attach an ideal ammeter to a battery, you will get infinite power.