Two capacitors of capacitances `3 muF` and `6 muF` are charged to a potential of `12 V` each. They are now connected to each other, with the positive plate of each joined to the negative plate of the other. The potential difference across each will be

Options

A. `6` volt
B. `4` volt
C. `3` volt
D. Zero

Explanation

To solve the problem step by step, we will analyze the given capacitors and their connections.

### Step 1: Calculate the charge on each capacitor
We have two capacitors:
- Capacitor 1 (C1) = 3 µF, charged to a potential (V1) = 12 V
- Capacitor 2 (C2) = 6 µF, charged to a potential (V2) = 12 V

The charge (Q) on a capacitor is given by the formula:
\[ Q = C \times V \]

For Capacitor 1:
\[ Q_1 = C_1 \times V_1 = 3 \, \mu F \times 12 \, V = 36 \, \mu C \]

For Capacitor 2:
\[ Q_2 = C_2 \times V_2 = 6 \, \mu F \times 12 \, V = 72 \, \mu C \]

### Step 2: Determine the total charge when connected
When the capacitors are connected with the positive plate of one connected to the negative plate of the other, the charges will combine. The total charge (Q_total) can be calculated as:
\[ Q_{\text{total}} = Q_2 - Q_1 = 72 \, \mu C - 36 \, \mu C = 36 \, \mu C \]

### Step 3: Calculate the equivalent capacitance
Since the capacitors are connected in a series-like manner (positive to negative), we need to find the equivalent capacitance (C_eq). The formula for equivalent capacitance in series is:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} \]

Calculating this:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{3 \, \mu F} + \frac{1}{6 \, \mu F} \]
\[ \frac{1}{C_{\text{eq}}} = \frac{2}{6 \, \mu F} + \frac{1}{6 \, \mu F} = \frac{3}{6 \, \mu F} = \frac{1}{2 \, \mu F} \]
Thus,
\[ C_{\text{eq}} = 2 \, \mu F \]

### Step 4: Calculate the potential difference across the equivalent capacitor
The potential difference (V) across the equivalent capacitor can be calculated using the formula:
\[ V = \frac{Q_{\text{total}}}{C_{\text{eq}}} \]

Substituting the values:
\[ V = \frac{36 \, \mu C}{2 \, \mu F} = 18 \, V \]

### Step 5: Determine the potential difference across each capacitor
Since the capacitors are connected in such a way that they share the same charge, the potential difference across each capacitor can be calculated as follows:

For Capacitor 1 (C1 = 3 µF):
\[ V_1 = \frac{Q_{\text{total}}}{C_1} = \frac{36 \, \mu C}{3 \, \mu F} = 12 \, V \]

For Capacitor 2 (C2 = 6 µF):
\[ V_2 = \frac{Q_{\text{total}}}{C_2} = \frac{36 \, \mu C}{6 \, \mu F} = 6 \, V \]

### Final Answer
The potential difference across Capacitor 1 is 12 V, and across Capacitor 2 is 6 V.

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