Question

The charge on a capacitor decreases $\eta$ times in time t, when it discharges through a circuit with a time constant $\tau$ .
(A) $t = \eta \tau$
(B) $t = \tau \ln \eta$
(C) $t = \tau (\ln \eta - 1)$
(D) $t = \tau \ln (1 - \dfrac{1}{\eta })$

Answer 1

Hint In this question, we are given that the charge on a condenser reduces by $\eta$, this means that if the original charge was q, the new charge would be $\dfrac{q}{\eta }$. We know that the charge in a capacitor reduces exponentially. So we need to find the final charge after time t and replace it by $\dfrac{q}{\eta }$

Complete step by step solution
A capacitor also known as a condenser is a device which can store a large amount of charge in a small space whereas capacitance is the ability of a device to store charge. When the resistance capacitor (R-C) circuit is connected to a power source (charging state) the equation of the circuit is given by
$\dfrac{q}{\eta }$
Where, E = Source e.m.f (electromotive force)
In case of discharging the power source is disconnected and the above equation becomes
$0 = RI + \dfrac{Q}{C}$
Or, $R\dfrac{{dQ}}{{dt}} + \dfrac{Q}{C} = 0$
Where, I = $\dfrac{{dQ}}{{dt}}$
When we integrate the above equation with the initial condition of t = 0, we get
$Q = {Q_0}{e^{ - \dfrac{t}{\tau }}}$ …..(i)
Where, Q = Charge on the capacitor at any instance of time.
${Q_0}$ = Initial/maximum charge on the capacitor.
t = Time.
$\tau$ = Time constant.
According to the question the charge on a capacitor decreases η times in time t
$\therefore$ Q = $\dfrac{{{Q_0}}}{\eta }$
Putting the value of Q in equation (i), we get
$\dfrac{{{Q_0}}}{\eta } = {Q_0}{e^{ - \dfrac{t}{\tau }}}$
Cancelling ${Q_0}$ from both sides, we get
$\dfrac{1}{\eta } = {e^{ - \dfrac{t}{\tau }}}$
Or, $\eta = {e^{\dfrac{t}{\tau }}}$
Taking log with base “e” on both sides, we get
$\dfrac{t}{\tau } = \ln \eta$

Or, $t = \tau \ln \eta$. (Option B)

Note Discharge of a capacitor always follows an exponential curve. The actual equation is represented by
$Q = {Q_0}{e^{ - \dfrac{t}{{RC}}}}$
Where, Q = Charge on the capacitor at any instance of time.
${Q_0}$ = Initial/maximum charge on the capacitor.
t = Time.
R = Resistance of the circuit.
C = Capacitance of the circuit.
Here,
$\dfrac{1}{{RC}}$ is called time constant and is represented by $\tau$. The above equation shows that the instantaneous charge on the capacitor decreases exponentially with time.