Book Reference : Pages To qualitatively understand how a capacitor discharges through a resistor 2.To derive the equation which defines this rate of discharge 3.To be able to solve capacitor discharge problems

When a charged capacitor is allowed to discharge through a fixed resistor it does so gradually until it reaches 0 V0V0 Switch ChargeDischarge 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 C Fixed resistor R

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 The first graph shows charge from Q=CV The second graph shows current from I = V/R [Virtual Physics Lab]

Consider charge, if we start at a charge of Q 0, then after a certain time t the charge will decay to say only 0.9Q 0 (arbitrary choice) Experimentally, we can show that after a further time t the charge has decayed to 0.9 x 0.9 Q 0 after 2 t and 0.9 x 0.9 x 0.9Q 0 after 3 t and 0.9 n Q 0 after time nt The decay is exponential

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/Rfrom Q=CV, V = Q/C so I = Q/CR

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 = Q 0 e –t/RC Where Q 0 is the initial charge & e is the exponential function the exponential function

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 : ChargeQ = Q 0 e –t/RC VoltageV = V 0 e –t/RC CurrentI = I 0 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

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

The initial charge on the capacitor Using Q=CV, the initial charge Q 0 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 = V 0 e -1 = 0.37 x 9V = 3.3V

The pd after 300s Using V = V 0 e –t/RC -t/RC = 300/220 = 1.36(no units) V = 9 e V = 2.3V

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  ]

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]