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 sec ?

Options

A. After 2ms from start, the potential difference across the resistor is 64 V
B. After 2ms from start, the potential difference across the capacitor is 64 V
C. After 2ms from start , the potential difference across the resistor is 60V
D. After 2ms from start , the potential difference across the capacitor is 60 V

Answer

To solve the problem, we need to understand the exponential decay of the potential difference (P.D.) across a capacitor when it is discharged through a resistor. The voltage across the capacitor can be described by the equation:

\[ V(t) = V_0 e^{-t/(RC)} \]

Where:
- \( V(t) \) is the voltage at time \( t \),
- \( V_0 \) is the initial voltage (100 V in this case),
- \( R \) is the resistance,
- \( C \) is the capacitance,
- \( t \) is the time.

### Step 1: Determine the decay constant \( \frac{1}{RC} \)

From the problem, we know that after 1 second, the voltage is 80 V. We can use this information to find the value of \( \frac{1}{RC} \).

Using the formula:

\[ V(1) = V_0 e^{-1/(RC)} \]

Substituting the known values:

\[ 80 = 100 e^{-1/(RC)} \]

Dividing both sides by 100:

\[ 0.8 = e^{-1/(RC)} \]

Taking the natural logarithm of both sides:

\[ \ln(0.8) = -\frac{1}{RC} \]

Thus,

\[ \frac{1}{RC} = -\ln(0.8) \]

### Step 2: Calculate the voltage after 2 seconds

Now we can find the voltage after 2 seconds using the same formula:

\[ V(2) = V_0 e^{-2/(RC)} \]

Substituting \( V_0 = 100 \) V:

\[ V(2) = 100 e^{-2/(RC)} \]

We already have \( e^{-1/(RC)} = 0.8 \), so:

\[ e^{-2/(RC)} = (e^{-1/(RC)})^2 = (0.8)^2 = 0.64 \]

Now substituting this back into the voltage equation:

\[ V(2) = 100 \times 0.64 = 64 \, \text{V} \]

### Final Answer

The potential difference between the plates of the capacitor after 2 seconds is **64 V**.

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