Course Updates http://www.phys.hawaii.edu/~varner/PHYS272-Spr10/physics272.html Notes for today: 1) Complete Chap 26, Problem session Friday 2)Assignment 6 (Mastering Physics) online and separate, written problems due Monday 3)Quiz 3 on Friday 4)Review Midterm 1
Kirchhoff’s Rules Kirchhoff’s rules are statements used to solve for currents and voltages in complicated circuits. The rules are Rule I. Sum of currents into any junction = ∑ is zero. I 0 i + = i I I I 1 2 12 Why? Since charge is conserved. a b Rule II. Sum of potential differences in any loop is zero. (This includes emfs) = ∑ V 0 i c i d + + + = Why? Since potential (energy) V V V V 0 ab bc cd da is conserved
RC Circuits I I I a a I R R b b + + C C ε ε - - RC 2 RC C ε RC 2 RC C ε − = C e t RC ε / q ( ) q − q = ε 1 − e t RC / q C 0 0 t t
RC Circuits (Time-varying currents, charging) I a I Charge capacitor: • R C initially uncharged; b connect switch to a at t =0 C ε Calculate current and charge as function of time. Would it matter where R Q • Loop theorem ⇒ ε − − = IR 0 is placed in the loop?? C • Convert to differential equation for Q : dQ I = dQ Q ⇒ ε = + R dt dt C
Charging Capacitor I a I Charge capacitor: • R dQ Q b ε = + R dt C C ε • Guess solution: − = ε − t RC Q C (1 e ) • Check that it is a solution: Note that this “guess” ⎛ ⎞ fits the boundary dQ 1 = ε − − t RC / ⎜ ⎟ C e conditions: ⎝ ⎠ dt RC = ⇒ = t 0 Q 0 − dQ Q t − + = − ε + ε − = ε t / RC ⇒ ! R e ( 1 e RC ) = ∞ ⇒ = ε t Q C dt C
Charging Capacitor I a I Charge capacitor: • R ( ) − = ε − t RC / b Q C 1 e C ε Current is found from • differentiation: ε ⇒ dQ Conclusion: = = − t RC / I e dt R Capacitor reaches its final • charge( Q = C ε ) exponentially with time constant τ = RC . Current decays from max • (= ε / R ) with same time constant.
Charging Capacitor RC 2 RC Charge on C C ε ( ) − = ε − t RC / Q C 1 e Max = C ε Q 63% Max at t = RC 0 t ε / R Current ε dQ − = = t RC / I e dt R I Max = ε / R 37% Max at t = RC 0 t
Discharging Capacitor I I a dQ Q + = R R dt 0 b C + + C ε - - • Guess solution: − τ − = = ε t / t RC / Q Q 0 e C e • Check that it is a solution: Note that this “guess” ⎛ ⎞ dQ 1 = ε − − t RC / fits the boundary ⎜ ⎟ C e ⎝ ⎠ conditions: dt RC = 0 ⇒ = ε t Q C − − ⇒ R dQ Q + = − ε + ε = = ∞ ⇒ = t RC / t RC / ! e e t Q 0 0 dt C
Discharging Capacitor I I • Discharge capacitor: a = − τ = ε − R t / t / RC Q Q 0 e C e b + + C ε - - • Current is found from differentiation: ε dQ ⇒ − = = − t RC / I e Conclusion: dt R • Capacitor discharges exponentially with time constant τ = RC Minus sign: Current is opposite to • Current decays from initial max original definition, value (= - ε / R ) with same time i.e., charges flow constant away from capacitor.
Discharging Capacitor RC 2 RC C ε Charge on C − = C ε t RC / Q e Max = C ε Q 37% Max at t = RC 0 t zero 0 Current ε I dQ = = − − t RC / I e dt R “Max” = - ε / R - ε / R 37% Max at t = RC t
A Midterm 1 statistics B C D F
Cumulative statistics
Midterm Review • Not many systematic, consistent problems • However, some recurrent themes: – Electric Potential (V) vs. Potential Energy (U) – Electric field sums as vector quantity – Potential field (V) sums as a scaler – Area of a circle is π r 2 • Detailed points breakdown (by roster ID) linked from course webpage • Will go over in outline form problems next…
For next time • HW #6 Assigned � due next Monday • Quiz #3 on Friday • Problem session Friday prior to quiz
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