Physics A2 32 CapacitorsDischargeThroughFixedResistor

8/13/2019 Physics A2 32 CapacitorsDischargeThroughFixedResistor

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Book Reference : Pages 98-101

1. 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

8/13/2019 Physics A2 32 CapacitorsDischargeThroughFixedResistor

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When a charged capacitor is allowed to discharge

through a fixed resistor it does so gradually untilit reaches 0

V0

Switch

Charge Discharge

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

CFixed

resistor R

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Look at the shape of the graphs qualitatively .

They both show curves which starts at the Y axisand decay asymptotically towards the X axis

The first graph

shows charge fromQ=CV

The second graph

shows current from

I = V/R

[Virtual Physics Lab]

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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 thecharge 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 0.9n Q 0 after time nt

The decay is exponential

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Exponential decays....

If a quantity decreases at a rate which is proportional tothe quantity (left) then the decay is exponential

Explaining the decrease

Consider one small step in the decay process where Qdecays 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

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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 ( t0)

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 

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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 = Q 0 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

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A 2200F 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

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The initial charge on the capacitor

Using Q=CV, the initial charge Q 0 is 2200F x 9V

= 0.02 C

The time constant for the circuit

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

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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

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A 50F capacitor is charged by connecting it to a 6V

battery & then discharging it through a 100k  resistor.

Calculate :

The initial charge stored [300C]

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 the100k  if 99% of the discharge is to be complete in about

5s [20k ] 

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A 68F capacitor is charged by connecting it to a 9V

battery & then discharging it through a 20k  resistor.

Calculate :

The initial charge stored [0.61C]

The initial discharge current [0.45mA]The pd and the discharge current 5s after the start

of the discharge [0.23V, 11A]