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

We have given that a capacitor is in series with a resistor and is charged to a potential of $100V$.

Thus,$V_0 =100V$ ,where $V_0$ is the potential difference or emf across the capacitor.

Let $C$ is the capacitance of the given capacitor and $R$ is the resistance of the given resistor.

Therefore, time constant is given by $\tau =RC$ .

Now, for discharging we know that voltage across the capacitor drops exponentially, which is given by

$$V=V_0 e^{-\frac{t}{RC}}\\ \implies V=V_0 e^{-\frac{t}{\tau}} \hspace{1cm}(\clubs)$$

Now, we have also given that potential drops across the capacitor in $t=1.2s$ is $V=67 \:Volt$

Thus from $(\clubs)$ we get,

$$67=100e^{-\frac{1.2}{\tau}}\\ \implies ln \bigg(\frac{67}{100}\bigg)=-\frac{1.2}{\tau}\\ \implies \tau=-\frac{1.2}{ ln \bigg(\frac{67}{100}\bigg)} \hspace{1cm}(\clubs \clubs)$$

Thus, on putting $(\clubs \clubs)$ in $(\clubs)$ we get,

$$V=V_0 e^{\frac{t}{\frac{1.2}{ ln \big(\frac{67}{100}\big)}}}\\ \implies V=V_0 e^{\frac{t }{1.2} ln \big(\frac{67}{100}\big)}\\ \implies V=V_0e^{ ln \big(\frac{67}{100}\big)^{\frac{t}{1.2}}}\\ \implies V=V_0 \big(\frac{67}{100}\big)^{\frac{t}{1.2}} \hspace{1cm} (\because e^{lnx}=x) \:(\spades)$$

Now from equation $(\spades)$ we can calculate at $t=t_1$ at where potential across the capacitor is $30V$ ,

$$30=100 \big(\frac{67}{100}\big)^{\frac{t_1}{1.2}}\\ \implies \frac{3}{10}=\big(\frac{67}{100}\big)^{\frac{t_1}{1.2}}\\ \implies ln \big(\frac{3}{10}\big)=\frac{t_1}{1.2}ln \big(\frac{67}{100}\big)\\ \implies t_1=1.2 \frac{ln(0.3)}{ln(0.67)}\\ \implies t_1=3.6s$$

Hence, required time will be $3.6s$ to reach $30V$ potential across the given capacitor.