Question

A capacitor is charged to a potential difference of 100 V and is then connected across a resistor. The potential difference across the capacitor decays exponentially with respect to time. After 1 sec, the P.D. between the plates of the capacitor is 80 V. What will be the potential difference between the plates after 2 seconds?

• A. 32 V
• B. 48 V
• C. 64 V
• D. 70 V

Answer 1

Answer: (C)

64 V

Answer 2

The potential difference across a capacitor discharging through a resistor decays exponentially with time, given by the formula:

$V(t) = V_0 e^{-t/\tau}$

where $V(t)$ is the potential difference at time $t$, $V_0$ is the initial potential difference, and $\tau = RC$ is the time constant of the circuit.

Given: Initial potential difference, $V_0 = 100$ V. At $t=1$ sec, the potential difference $V(1) = 80$ V.

Using the formula for $t=1$ sec: $80 = 100 e^{-1/\tau}$ Divide both sides by 100: $e^{-1/\tau} = \frac{80}{100} = \frac{4}{5}$

Now, we need to find the potential difference after $t=2$ seconds, $V(2)$. Using the formula for $t=2$ sec: $V(2) = V_0 e^{-2/\tau}$ We can rewrite $e^{-2/\tau}$ as $(e^{-1/\tau})^2$: $V(2) = V_0 (e^{-1/\tau})^2$ Substitute the values $V_0 = 100$ V and $e^{-1/\tau} = \frac{4}{5}$: $V(2) = 100 \left(\frac{4}{5}\right)^2$ $V(2) = 100 \times \frac{16}{25}$ $V(2) = 4 \times 16$ $V(2) = 64$ V

The potential difference between the plates after 2 seconds will be 64 V.