Two capacitors of capacitance `2muF and 5muF` are charged to a potential difference 100V and 50 V respectively and connected such that the positive plate of one capacitor is connected to the negative plate of the other capacitor after the switch is closed, the initial current in the circuit is 50 mA. the total resistance of the connecting wires is (in Ohm):

Options

A. 100
B. 300
C. 1000
D. 3000

Explanation

To solve the problem step by step, we will analyze the given information and apply Kirchhoff's Voltage Law (KVL) to find the total resistance of the connecting wires.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Capacitor 1 (C1) = 2 µF, charged to V1 = 100 V
- Capacitor 2 (C2) = 5 µF, charged to V2 = 50 V
- Initial current (I) = 50 mA = 50 × 10^(-3) A

2. **Determine the Voltage Difference:**
- When the positive plate of C1 is connected to the negative plate of C2, the effective voltage across the circuit can be calculated as:
\[ V_{\text{total}} = V1 + V2 = 100 \, \text{V} + 50 \, \text{V} = 150 \, \text{V} \]

3. **Apply Kirchhoff's Voltage Law (KVL):**
- According to KVL, the sum of the potential differences in a closed loop is equal to zero. For our circuit, we can write:
\[ V_{\text{total}} - I \cdot R = 0 \]
- Rearranging gives:
\[ V_{\text{total}} = I \cdot R \]

4. **Substitute Known Values:**
- Substitute the values of \(V_{\text{total}}\) and \(I\):
\[ 150 \, \text{V} = (50 \times 10^{-3} \, \text{A}) \cdot R \]

5. **Solve for Resistance (R):**
- Rearranging the equation to find R:
\[ R = \frac{150 \, \text{V}}{50 \times 10^{-3} \, \text{A}} = \frac{150}{0.05} = 3000 \, \Omega \]

### Final Answer:
The total resistance of the connecting wires is \( R = 3000 \, \Omega \).

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