Discharging Capacitors Through Resistors

1. Learning Objectives Book Reference : Pages 98-101 Discharging Capacitors Through Resistors To qualitatively understand how a capacitor discharges through a resistor To derive the equation which defines this rate of discharge To be able to solve capacitor discharge problems

2. When a charged capacitor is allowed to discharge through a fixed resistor it does so gradually until it reaches 0 Capacitor Discharge 1 Charge Discharge Switch C Fixed resistor R V0 • We can compare this discharge with water leaving a tank through a pipe at the bottom, initially the flow rate is high because of the pressure. At the level falls so does the pressure reducing the flow rate

3. Look at the shape of the graphs qualitatively. They both show curves which starts at the Y axis and decay asymptotically towards the X axis Capacitor Discharge 2 The first graph shows charge from Q=CV The second graph shows current from I = V/R [Virtual Physics Lab]

4. Consider charge, if we start at a charge of Q0, then after a certain time t the charge will decay to say only 0.9Q0 (arbitrary choice) • Experimentally, we can show that after a further time t the charge has decayed to • 0.9 x 0.9 Q0 after 2 t • and 0.9 x 0.9 x 0.9Q0 after 3 t • and 0.9n Q0 after time nt • The decay is exponential Capacitor Discharge 3

5. Exponential decays.... • If a quantity decreases at a rate which is proportional to the quantity (left) then the decay is exponential • Explaining the decrease • Consider one small step in the decay process where Q decays to Q - Q in a time t • The current at this time is given by • I = V/R from Q=CV, V = Q/C so • I = Q/CR Capacitor Discharge 4

6. From I = Q/t if t is very small then the drop in charge -Q can be rewritten as -It and I is therefore -Q/t • Substituting into our earlier equation for I = Q/CR • Q/t = -Q/CR • For infinitely short time intervals as t tends to 0 (t0) • Q/t represents the rate of change of charge & is written as the first differential dQ/dt hence • dQ/dt = -Q/CR • Solution by integration : • Q = Q0 e–t/RCWhere Q0 is the initial charge & e is • the exponential function Capacitor Discharge 4

7. From Q=CV voltage is proportional to charge, similarly from Ohm’s law Current is proportional to voltage. • All three quantities decay in exactly the same way : • Charge Q = Q0 e–t/RC • Voltage V = V0 e–t/RC • Current I = I0 e–t/RC • The quantity “RC” is called the time constant & is the time for the initial charge/voltage/current to fall to 0.37 of the initial value (0.37 = e-1) • The units for RC are the second Notes 1

8. A 2200F capacitor is charged to a pd of 9V and then allowed to discharge through a 100k resistor. Calculate • The initial charge on the capacitor • The time constant for the circuit • The pd after a time equal to the time constant • The pd after 300s Worked Example

9. The initial charge on the capacitor • Using Q=CV, the initial charge Q0 is 2200F x 9V • = 0.02 C • The time constant for the circuit • Time constant = RC = 100,000 x 2200F = 220s • The pd after a time equal to the time constant • By definition t = RC when V = V0e-1 = 0.37 x 9V = 3.3V Worked Example

10. The pd after 300s • Using V = V0 e–t/RC • -t/RC = 300/220 = 1.36 (no units) • V = 9 e-1.36 • V = 2.3V Worked Example

11. A 50F capacitor is charged by connecting it to a 6V battery & then discharging it through a 100k resistor. Calculate : • The initial charge stored [300C] • The time constant for the circuit [5.0s] • Estimate how long the capacitor would take to discharge to about 2V [5s] • Estimate the size of the resistor required in place of the 100k if 99% of the discharge is to be complete in about 5s [20k] Problems 1

12. A 68F capacitor is charged by connecting it to a 9V battery & then discharging it through a 20k resistor. Calculate : • The initial charge stored [0.61C] • The initial discharge current [0.45mA] • The pd and the discharge current 5s after the start of the discharge [0.23V, 11A] Problems 2