RC Circuits - circuits in which the currents vary in time - rate of charging a cap depends on C and R of circuit - differential equations

Discharging a Capacitor (- sign because q decreases for I > 0 That is, current in circuit equals the decrease of charge on the capacitor) C R q -q I Given: R, C, q o (initial charge) Find: q(t) and I(t) when switch is closed 2) 1) (Kirchhoff’s Loop Rule)

C R q -q I where: q = q(t) q(0) = q o This is a differential equation for the function q(t), subject to the initial condition q(0) = q 0. We are looking for a function which is proportional to its own first derivative (since dq/dt ~ -q). Combine 1) and 2) to get:

RC is called the “time constant” or “characteristic time” of the circuit. Units: 1 Ω x 1 F = 1 second (show this!) Write  (“tau”) = RC, then: (discharging) Solution:

Discharging q qoqo 23 t t =, q ≈ 0.37 q o = (q o /e) t = 2, q ≈ 0.14 q o = (q o /e 2 ) t = 3, q ≈ 0.05 q o = (q o /e 3 ) t  ∞, q  0 = (q o /e ∞ )

Draw a graph for I(t).

Example 1 A capacitor is charged up to 18 volts, and then connected across a resistor. After 10 seconds, the capacitor voltage has fallen to 12 volts. a)What is the time constant RC ? b)What will the voltage be after another 10 seconds (20 seconds total)? A)8V B)6V C)4V D)0

Charging a capacitor C is initially uncharged, and the switch is closed at t=0. After a long time, the capacitor has charge Q f. R C Then, where  RC. Question: What is Q f equal to? What is V(t) ? Recall, C=Q/V, so V(t)=Q(t)/C

Charging a capacitor t = 0, q=0t = 3 RC, q  0.95 Q f t = RC, q  0.63 Q f etc. t = 2 RC, q  0.86 Q f q QfQf 23 t

Draw a graph of I(t). Why is I=+dq/dt this time?

Example kΩ 12 V 2 µF i)Initial current ii) Initial voltage across the resistor iii) Initial voltage across the capacitor iv)Time for voltage across C to reach 0.63*12V v) Final voltage across the resistor vi) Final voltage across the capacitor The capacitor is initially uncharged. After the switch is closed, find:

Solution

Example 3 The circuit below contains two resistors, R1 = 2.00 kΩ and R2 = 3.00 kΩ, and two capacitors, C1 = 2.00 μF and C2 = 3.00 μF, connected to a battery with emf ε = 120 V. No charge is on either capacitor before switch S is closed. Determine the charges q1 and q2 on capacitors C1 and C2, respectively, after the switch is closed.

Solution: 1) reconstruct the circuit so that it becomes a simple RC circuit containing a single resistor and single capacitor 2) determine the total charge q stored in the equivalent circuit.

“RC” Circuits a capacitor takes time to charge or discharge through a resistor “time constant” or “characteristic time” = RC (1 ohm) x (1 farad) = 1 second