Book Reference : Pages 98-101Book Reference : Pages 98-101

1.1. To qualitatively understand how a capacitor To qualitatively understand how a capacitor discharges through a resistordischarges through a resistor

2.2. To derive the equation which defines this To derive the equation which defines this rate of dischargerate of discharge

3.3. To be able to solve capacitor discharge To be able to solve capacitor discharge problemsproblems

When a charged capacitor is allowed to discharge When a charged capacitor is allowed to discharge through a fixed resistor it does so gradually until through a fixed resistor it does so gradually until it reaches 0it reaches 0

V0

Switch

Charge Discharge

We can compare this discharge with water leaving a We can compare this discharge with water leaving a tank through a pipe at the bottom, initially the flow tank through a pipe at the bottom, initially the flow rate is high because of the pressure. At the level falls so rate is high because of the pressure. At the level falls so does the pressure reducing the flow ratedoes the pressure reducing the flow rate

C Fixed resistor R

Look at the shape of the graphs Look at the shape of the graphs qualitativelyqualitatively. . They both show curves which starts at the Y axis They both show curves which starts at the Y axis and decay asymptotically towards the X axisand decay asymptotically towards the X axis

The first graph The first graph shows charge from shows charge from Q=CVQ=CV

The second graph The second graph shows current from shows current from I = V/RI = V/R

[Virtual Physics Lab]

Consider charge, if we start at a Consider charge, if we start at a charge of Qcharge of Q00, then after a certain , then after a certain time t the charge will decay to time t the charge will decay to say only 0.9Qsay only 0.9Q0 0 (arbitrary choice)(arbitrary choice)

Experimentally, we can show Experimentally, we can show that after a further time t the that after a further time t the charge has decayed tocharge has decayed to

0.9 x 0.9 Q0.9 x 0.9 Q00 after 2 t after 2 t

and 0.9 x 0.9 x 0.9Qand 0.9 x 0.9 x 0.9Q00 after 3 t after 3 t

and 0.9and 0.9nn Q Q00 after time nt after time nt

The decay is exponential The decay is exponential

From I = Q/t if From I = Q/t if t is very small then the drop in charge -t is very small then the drop in charge -Q can be rewritten as -IQ can be rewritten as -It and I is therefore -t and I is therefore -Q/Q/tt

Substituting into our earlier equation for I = Q/CRSubstituting into our earlier equation for I = Q/CR

Q/Q/t = -Q/CRt = -Q/CR

For infinitely short time intervals as For infinitely short time intervals as t tends to 0 (t tends to 0 (tt0)0)

Q/Q/t represents the rate of change of charge & is t represents the rate of change of charge & is written as the first differential dQ/dt hencewritten as the first differential dQ/dt hence

dQ/dt = -Q/CRdQ/dt = -Q/CR

Solution by integration :Solution by integration :

Q = QQ = Q00 e e–t/RC–t/RC Where QWhere Q00 is the initial charge & e is is the initial charge & e is the exponential functionthe exponential function

A 2200A 2200F capacitor is charged to a pd of 9V and then F capacitor is charged to a pd of 9V and then allowed to discharge through a 100kallowed to discharge through a 100k resistor. Calculate resistor. Calculate

The initial charge on the capacitorThe initial charge on the capacitor

The time constant for the circuitThe time constant for the circuit

The pd after a time equal to the time constantThe pd after a time equal to the time constant

The pd after 300sThe pd after 300s

The initial charge on the capacitorThe initial charge on the capacitor

Using Q=CV, the initial charge QUsing Q=CV, the initial charge Q00 is 2200 is 2200F x 9V F x 9V

= 0.02 C= 0.02 C

The time constant for the circuitThe time constant for the circuit

Time constant = RC = 100,000Time constant = RC = 100,000 x 2200 x 2200F = F = 220s220s

The pd after a time equal to the time constantThe pd after a time equal to the time constant

By definition t = RC when V = VBy definition t = RC when V = V00ee-1-1 = 0.37 x 9V = = 0.37 x 9V = 3.3V3.3V

The pd after 300sThe pd after 300s

Using V = VUsing V = V00 e e–t/RC–t/RC

-t/RC = 300/220 = 1.36-t/RC = 300/220 = 1.36 (no units)(no units)

V = 9 eV = 9 e-1.36-1.36

V = 2.3V V = 2.3V

A 50A 50F capacitor is charged by connecting it to a 6V F capacitor is charged by connecting it to a 6V battery & then discharging it through a 100kbattery & then discharging it through a 100k resistor. resistor. Calculate :Calculate :

The initial charge stored [300The initial charge stored [300C]C]

The time constant for the circuit [5.0s]The time constant for the circuit [5.0s]

Estimate how long the capacitor would take to discharge Estimate how long the capacitor would take to discharge to about 2V [5s]to about 2V [5s]

Estimate the size of the resistor required in place of the Estimate the size of the resistor required in place of the 100k100k if 99% of the discharge is to be complete in about if 99% of the discharge is to be complete in about 5s [5s [20k20k]]

A 68A 68F capacitor is charged by connecting it to a 9V F capacitor is charged by connecting it to a 9V battery & then discharging it through a 20kbattery & then discharging it through a 20k resistor. resistor. Calculate :Calculate :

The initial charge stored [0.61The initial charge stored [0.61C]C]

The initial discharge current [0.45mA]The initial discharge current [0.45mA]

The pd and the discharge current 5s after the start The pd and the discharge current 5s after the start of the discharge [0.23V, 11of the discharge [0.23V, 11A]A]