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 situation involving the two capacitors and apply Kirchhoff's Voltage Law (KVL) to find the total resistance in the circuit.

### Step 1: Understand the Capacitors and Their Voltages
We have two capacitors:
- Capacitor 1 (C1): 2 µF charged to 100 V
- Capacitor 2 (C2): 5 µF charged to 50 V

### Step 2: Determine the Initial Voltage Across the Circuit
When the capacitors are connected such that the positive plate of C1 is connected to the negative plate of C2, the total voltage across the circuit will be the difference in their voltages:
- Voltage across C1 = 100 V (positive terminal)
- Voltage across C2 = 50 V (negative terminal)

The effective voltage (V_total) in the circuit after connecting the capacitors is:
\[ V_{\text{total}} = V_{C1} - V_{C2} = 100 \, \text{V} - 50 \, \text{V} = 50 \, \text{V} \]

### Step 3: Use Kirchhoff's Voltage Law (KVL)
According to KVL, the sum of the potential differences in a closed loop is equal to the sum of the voltage drops across the resistances. In this case, we have:
\[ V_{\text{total}} = I \cdot R \]
Where:
- \( V_{\text{total}} = 50 \, \text{V} \)
- \( I = 50 \, \text{mA} = 50 \times 10^{-3} \, \text{A} \)
- \( R \) is the total resistance we need to find.

### Step 4: Substitute the Known Values
Substituting the known values into the KVL equation:
\[ 50 \, \text{V} = (50 \times 10^{-3} \, \text{A}) \cdot R \]

### Step 5: Solve for Resistance (R)
Rearranging the equation to solve for \( R \):
\[ R = \frac{50 \, \text{V}}{50 \times 10^{-3} \, \text{A}} \]
\[ R = \frac{50}{0.05} \]
\[ R = 1000 \, \Omega \]

### Step 6: Conclusion
The total resistance of the connecting wires is:
\[ R = 1000 \, \Omega \]