Answer to Question #116842 in Molecular Physics | Thermodynamics for Phoebe

Question

A capacitor is in series with a resistor and is charged to a potential of 100 V. When discharging the capacitor it takes 1.2 s for the potential across the capacitor to drop to 67 V. How long does it take the potential difference across the capacitor to drop to 30 V? (to 2 s.f and in s)

Answer 1

As per the question,

Potential = 100V

time of discharging the capacitor = 1.2 sec

Potential developed across the capacitor =67V

Time taken by the capacitor to drop the potential across the capacitor till 30V,

Let the initial charge was Q,

Let the resistance of the circuit is R,

Applying the KVL rule,

$\Rightarrow IR+\dfrac{Q}{c}=0$

$\Rightarrow \dfrac{dQ}{dt}R+\dfrac{Q}{c}=0$

$\Rightarrow \dfrac{dQ}{Q}=-\dfrac{1}{RC}dt$

Now, taking the integration of both side,

$\int_{Q1}^{Q_2} \dfrac{dQ}{Q}=-\dfrac{1}{RC}\int_0^t dt$

$\Rightarrow \ln Q_2-\ln Q_1=\dfrac{-t}{RC}$

$\Rightarrow Q_2=Q_1e^{-\frac{-t}{RC}}$

Dividing both side w ith C,

$\dfrac{V_2}{V_1}=e^{\frac{-t}{RC}}$

Now, taking log of both side,

$\Rightarrow \ln(\dfrac{V_2}{V_1})=\dfrac{-t}{RC}$

$\Rightarrow \ln(\dfrac{67}{100})=\dfrac{-1.2}{RC}$

$\Rightarrow RC=\dfrac{-1.2}{\ln (67/100)}$

Now, again, applying the same,

$\ln(\dfrac{30}{100})=\dfrac{t\ln(67/100)}{1.2}$

$\Rightarrow t=1.2\ln (\dfrac{30}{100})\dfrac{1}{\ln(67/100)}$

$t=3.60sec$