The plates of a capacitor are charged to a potential difference of 320 volt and are then connected across a resistor. The potential difference across the capacitor decays exponentially with time. After 1 sec the potential difference between the plates of the capacitor is 240 volts. what is the potential difference between the plates after 2 s?

Options

A. 200 V
B. 180 V
C. 160 V
D. 140 V

Explanation

To solve the problem step by step, we can follow these calculations:

### Step 1: Understand the Exponential Decay of Voltage
The voltage across a capacitor discharges exponentially when connected to a resistor. The voltage at any time \( t \) can be expressed as:
\[ V(t) = V_0 e^{-\lambda t} \]
where:
- \( V(t) \) is the voltage at time \( t \),
- \( V_0 \) is the initial voltage (320 V in this case),
- \( \lambda \) is the decay constant,
- \( t \) is the time in seconds.

### Step 2: Use Given Information to Find the Decay Constant
We know that after 1 second, the voltage \( V(1) = 240 \) V. Plugging this into the equation gives:
\[ 240 = 320 e^{-\lambda \cdot 1} \]
To isolate \( e^{-\lambda} \), we divide both sides by 320:
\[ e^{-\lambda} = \frac{240}{320} = \frac{3}{4} \]

### Step 3: Find \( e^{-\lambda} \) for \( t = 2 \) seconds
Now, we need to find the voltage after 2 seconds. We can express this as:
\[ V(2) = V_0 e^{-\lambda \cdot 2} \]
Since we have \( e^{-\lambda} = \frac{3}{4} \), we can find \( e^{-2\lambda} \) as follows:
\[ e^{-2\lambda} = (e^{-\lambda})^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16} \]

### Step 4: Calculate the Voltage at \( t = 2 \) seconds
Now substituting back into the voltage equation:
\[ V(2) = 320 \cdot e^{-2\lambda} = 320 \cdot \frac{9}{16} \]
Calculating this gives:
\[ V(2) = 320 \cdot \frac{9}{16} = 20 \cdot 9 = 180 \text{ volts} \]

### Conclusion
Thus, the potential difference between the plates of the capacitor after 2 seconds is **180 volts**.

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