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Q. A capacitor of capacitance $C$ is charged to a potential difference $V$ and then connected in series with an open key and a pure resistor $R$ . At time $t=0$ , the key is closed. If $I$ is current at time $t=0$ , a plot of $log I$ against $t$ is shown in the graph $\left(\right.2\left.\right)$ . Later one of the parameters, i.e., $V$ , $R$ and $C$ is changed, keeping the other two constant and graph $\left(\right.2\left.\right)$ is recorded. Then

Question

NTA AbhyasNTA Abhyas 2020Current Electricity

Solution:

During discharging of a capacitor, the current is given by,
$I=\frac{E}{R}e^{- \, \frac{t}{R C}}$
$\log I=\log ⁡ \frac{E}{R}e^{- \, \frac{t}{R C}}$
$\log I=-\frac{t}{R C}+\log ⁡ \frac{E}{R}$
Intercept is constant $\Rightarrow $ $E$ & $R$ constant
Magnitude of slope decreases $\Rightarrow $ $C$ is increased