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:

�

E +VR (t) +VC (t) = 0E − i(t)R − q(t) /C = 0E − dq /dt( )R − q(t) /C = 0

CE

VC=Q/C VR=iR

�

q(t) = CE 1− e− t /RC( )→ i(t) ≡ dq /dt = E /R( )e−t /RC

i(t) E/R

ττ

Time constant: τ = 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 + -

+++ ---

�

VR +VC = 0−i(t)R + q(t) /C = 0 → dq /dt( )R + q(t) /C = 0

Solution:

�

q(t) = q(0)e− t /RC = CEe−t /RC

i(t) = dq /dt = q(0) /RC( )e−t /RC = E /R( )e−t /RC

τ τ

i(t) E/R

τ

Time constant: τ = RC Time i drops to 1/e≅1/2.

(a) Compare i(0) = E / R

(b) Compare τ = RC

i1> i2 > i4 > i3

τ 4 > τ1 = τ 2 > τ 3

τ1 = 6, τ 2 = 6, τ 3 = 5, τ 4 = 10

i1 = 6, i2 = 4, i3 = 1, i4 = 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 Req. Then apply charging formula with ReqC = τ If capacitors between resistors in parallel use loop rule.

Example In the figure, E = 1 kV, C = 10 µF, R1 = R2 = R3 = 1 MΩ. 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. At t>>0 replace capacitor with break in wire, i3=0, and use loop rule on R1 and R2 branch only. A long time is t>>𝜏.

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:

�

E +VR (t) +VC (t) = 0E − i(t)R − q(t) /C = 0E − dq /dt( )R − q(t) /C = 0

CE

VC=Q/C VR=iR

�

q(t) = CE 1− e− t /RC( )→ i(t) ≡ dq /dt = E /R( )e−t /RC

i(t) E/R

ττ

Time constant: τ = 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 + -

+++ ---

�

VR +VC = 0−i(t)R + q(t) /C = 0 → dq /dt( )R + q(t) /C = 0

Solution:

�

q(t) = q(0)e− t /RC = CEe−t /RC

i(t) = dq /dt = q(0) /RC( )e−t /RC = E /R( )e−t /RC

τ τ

•  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 Req=∑Rj 1/Ceq= ∑1/Cj

In parallel: same voltage same voltage 1/Req= ∑1/Rj Ceq=∑Cj

P = iV = i2R = V2/R U = QV/2 = Q2/2C = CV2

Many Batteries? Loop & Junction!

One Battery: Simplify First Three Batteries: Straight to

Loop & Junction

1R23

par =1R2

+ 1R3

→ R23par = 12Ω

Too Many Batteries!