Lecture 21: MON 12 OCT Circuits III Physics 2113 Jonathan Dowling

RC Circuits: Charging a Capacitor In these circuits, current will change for a while, and then stay constant. We want to solve for current as a function of time i(t)=dq/dt. The charge on the capacitor will also be a function of time: q(t). The voltage across the resistor and the capacitor also change with time. To charge the capacitor, close the switch on a. A differential equation for q(t)! The solution is: CECE V C =Q/C V R =iR i(t) E /R Time constant:  = RC Time i drops to 1/e.

i(t) E /R RC Circuits: Discharging a Capacitor Assume the switch has been closed on a for a long time: the capacitor will be charged with Q=CE. Then, close the switch on b: charges find their way across the circuit, establishing a current. C +  +++  Solution:

i(t) E /R Time constant:  = RC Time i drops to 1/e ≅ 1/2.

Example The three circuits below are connected to the same ideal battery with emf E. All resistors have resistance R, and all capacitors have capacitance C. Which capacitor takes the longest in getting charged? Which capacitor ends up with the largest charge? What ’ s the final current delivered by each battery? What happens when we disconnect the battery? Simplify R ’ s into into R eq. Then apply charging formula with R eq C =    

Example In the figure, E = 1 kV, C = 10 µF, R 1 = R 2 = R 3 = 1 M. With C completely uncharged, switch S is suddenly closed (at t = 0). What ’ s the current through each resistor at t=0? What ’ s the current through each resistor after a long time? How long is a long time? At t=0 replace capacitor with solid wire (open circuit) then use loop rule or Req series/parallel.          

RC Circuits: Charging a Capacitor In these circuits, current will change for a while, and then stay constant. We want to solve for current as a function of time i(t)=dq/dt. The charge on the capacitor will also be a function of time: q(t). The voltage across the resistor and the capacitor also change with time. To charge the capacitor, close the switch on a. A differential equation for q(t)! The solution is: CECE V C =Q/C V R =iR i(t) E /R Time constant:  = RC Time i drops to 1/e.

i(t) E /R RC Circuits: Discharging a Capacitor Assume the switch has been closed on a for a long time: the capacitor will be charged with Q=CE. Then, close the switch on b: charges find their way across the circuit, establishing a current. C +  +++  Solution:

Fire Hazard: Filling gas can in pickup truck with plastic bed liner. Safe Practice: Always place gas can on ground before refueling. Touch can with gas dispenser nozzle before removing can lid. Keep gas dispenser nozzle in contact with can inlet when filling.

One Battery? Simplify! Resistors Capacitors Key formula: V=iR Q=CV In series: same current dQ/dt same charge Q R eq =∑R j 1/C eq = ∑1/C j In parallel: same voltage same voltage 1/R eq = ∑1/R j C eq =∑C j P = iV = i 2 R = V 2 /R U = QV/2 = Q 2 /2C = CV 2

Many Batteries? Loop & Junction! One Battery: Simplify First Three Batteries: Straight to Loop & Junction

Too Many Batteries!