RC−Circuit
Current Electricity of Class 12
RC−CIRCUIT
Charging
Let us assume that the capacitor in the shown network is uncharged for t < 0. The switch is connected to position 1 at t = 0. Now, 'C' is getting charged.
If the charge on capacitor at time 't' is q, writing the loop rule, q/c+ IR − E = 0


Integrating,
⇒q = EC[1 − e−t/RC]
⇒At t = 0, q = 0and at t = ∞, q = EC (the maximum charge.)


Time Constant (τ)
It is the time during which the charging would have been completed, had the growth rate been as it began initially. Numerically, it is equal to RC. The unit of time constant is sec., if resistance R is in ohm and capacitance C is in farad.
Illustration 30: A capacitor is connected to a 12 V battery through a resistance of 10 Ω. It is found that the potential difference across the capacitor rises to 4.0 V in 1 μs. Find the capacitance of the capacitor.
Solution: The charge on the capacitor during charging is given by
Q = Q0(1 − e−t/RC).
Hence, the potential difference across the capacitor is
V = Q/C = Q0/C(1−e−t/RC)
Here at t = 1 μs, the potential difference is 4V where as the steady state potential difference is Q0/C = 12 V. So,
4V = 12V(1 − e−t/RC)
or,1 − e−t/RC = 1/3
or, e−t/RC = 2/3
or, t/RC = ln(3/2)= 0.405
or, RC = t/0.405 = 1μs/0.405 = 2.469μS
or, C = 2.469μs/10Ω = 0.25 μF
Discharging
Consider the same arrangement that we had in previous case with one difference that the capacitor has charge qo for t < 0 and the switch is connected to position 2 at t = 0. If the charge on capacitor is q at any later moment t, then the loop equation is given as


Integrating, at t = 0, q = q0
t = t, q = q
'−ve' sign indicates that the discharging current flows in a direction opposite to the charging current.
Illustration 31: A capacitor charged to a 50 V is discharged by connecting the two plates through a resistor R at t = 0. If the potential difference across the plates drops to 1.0 V at t = 10 ms, what will be the potential difference at t = 20 ms?
Solution: The potential difference at time t is given by
V = Q/C = (Q0/C)e−t/RC
or, V = V0e−t/RC
According to the given data,
1V = (50 V)e−10 ms/RC
or, e−10 ms/RC = 1/50
The potential difference at t = 20 ms is
V = V0e−t/RC
= (50 V)e−20 ms/RC = (50 V)(e^{10ms/RC})^{2}
= 0.02 V