The growth of charge *Q* on a capacitor connected in series with a resistance* R* ohm and a direct voltage source of emf *V*_{0} volt is given by

*Q **= Q*_{0}*(1*** – e^{–t/RC}) **where

*Q*

_{0}=

*CV*

_{0}which is the final (maximum) charge on the capacitor and

*e*is the base of natural logarithms.

_{0}is allowed to discharge through a resistance R, the decay of charge on the capacitor is given by

*Q = Q*_{0}*e ^{–t/RC}*

The nature of the growth and the decay of charge are shown in the adjoining figure.

Since the voltage (*V*) across the capacitor is directly proportional to the charge on it, the

*V **= V*_{0}*(***1**** – e^{–t/RC}) **and

*V **= V*_{0}** e^{–t/RC} **where

*V*

_{0}is the charging battery

The ** time constant** of the CR circuit is

*RC*which is the time taken for the charge to grow from zero value to (1– 1/

*e*) times (0.632 times) the final maximum charge. This follows from the equation for the exponential growth of charge.

*RC*is also equal to the time taken for the charge to decay from the initial value to 1/

*e*times (0.368 times) the initialal value. This follows from the equation for the exponential decay of charge.

The charging current is obtained by differentiating the equation *Q **= Q*_{0}*(*1*– e ^{–t/RC})* with respect to the time

*t*. Thus the charging current

*(*

*dQ/dt*) is given by

*I* = (*Q _{0} /RC*)

*e*

^{–t/RC}= I_{0}

*e*where

^{–t/RC}*I*

_{0}=

*Q*=

_{0}/RC*V*

_{0}

*/R*where

*V*

_{0}is the charging

The exponential *decay* of the charging current is shown in the adjoining figure.

The current during the discharge of the capacitor through the resistor is obtained by differentiating the equation *Q **= Q*_{0}*e ^{–t/RC}* with respect to the time

*t*. Thus the discharging current

*(*

*dQ/dt*) is given by

*I* = – (*Q _{0} /RC*)

*e*–

^{–t/RC}=*I*

_{0}

*e*where

^{–t/RC}*I*

_{0}=

*Q*=

_{0}/RC*V*

_{0}

*/R*where

*V*

_{0}is the initial voltage on the capacitor.

The exponential *decay* of the discharging current which flows in the direction *opposite* to the charging current also is shown in the adjoining figure.

[The charging and discharging of a capacitor is similar to inflating and deflating a balloon. The air pressure is the analogue of

We will now discuss a few multiple choice questions in this section:

**(1) **A transistor time delay relay** **circuit makes use of a series RC circuit. If the circuit uses a 60 μF capacitor what is the resistance required to produce a

(a) 60 Ω

(b) 6×10^{–5} Ω

(c) 1.67×10^{4} Ω

(d) 10^{5} Ω

(e) 10^{6} Ω

The *RC = *60* *so that* *the resistance required is given by

*R = *60 s/60×10^{–6} F = 10^{6} Ω

**(2)** A constant current source charges an initially uncharged capacitor. During the charging process, the potential difference *V *between the plates of the capacitor is related to the charging time *t* as

(a) *V* α *t** *

(b) *V* α *t*^{0}* *

(c) *V* α *t*^{–1}

(d)^{ }*V* α (1– *e ^{–t}*)

(e)* V* α *e ^{–t}*

Since the current *I *is constant, the charge *Q = It*) is directly proportional to the time *t*. Since the potential difference *V *between the plates of the capacitor is directly proportional to the charge (*V = Q/C*), we have *V* α *t*.* *

**(3) **A 10 μF capacitor is connected in series with a 5 Ω resistor and a battery of negligible internal resistance and emf 6 V. After a long time, if the charging battery is removed and the capacitor is immediately connected across the 5 Ω resistor, the initial value of the discharge current is

(a) 1.2 A

(b) 0.6 A

(c) 0.3 A

(d) 0.83 A

(e) zero

The capacitor will get charged to the battery *initial* value of the discharge current (at *t = *0) is given by

*I* = –*I*_{0}* e ^{–t/RC} *where

*I*

_{0}=

*V*

_{0}

*/R*where

*V*

_{0}is the initial voltage on the capacitor. Let us ignore the negative sign which just indicates that the discharge current is opposite to the charging current.

Therefore, *I* = (6/5)* e*^{0} = 1.2 A

**(4) **In the above question what will be the

(a) 1.333 V

(b) 2.208 V

(c) 3.792 V

(d) 4.5 V

(e) 6 V

The time constant of the circuit is *RC =*5×10×10^{–6} = 50 microsecond.

After one time constant, the voltage on the capacitor will be (1– 1/*e*) times (0.632 times) the charging battery

**(5) **In question No.3 what will be the charging current at the instant *t* = 50 microseconds?

(a) 0.3680 A

(b) 0.4416 A

(c) 0.6 A

(d) 0.7326 A

(e) 1.2 A

Since the time constant of the circuit, *RC =*5×10×10^{–6} s = 50 microseconds, the charging current at the instant *t* = 50 microseconds will be *I* = *I*_{0}* e ^{–t/RC}* =

*I*

_{0}

*e*

^{–}^{1}= 0.368

*I*

_{0}.

But *I*_{0}* = V*_{0}*/R * = 6/5 = 1.2 A.

Therefore *I* = 0.368×1.2 = 0.4416 A.