Two capacitors of capacitance `2muF` and `3muF` respectively are charged to potential difference 20 V and 40 V respectively. Now the capacitors are connected in series with a resistance such that the positively charged plate of one capacitor is connected to the positively charged plate of the other. The initial current through the resistance is `I_(0)`. The potential difference across the `2muF` capacitor at the instant the current has reduced to `(I_(0))/(2)` is_______ V.

Explanation

To solve the problem, we need to find the potential difference across the 2μF capacitor when the current has reduced to \( \frac{I_0}{2} \).

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Capacitor \( C_1 = 2 \mu F \) charged to \( V_1 = 20 V \).
- Capacitor \( C_2 = 3 \mu F \) charged to \( V_2 = 40 V \).

2. **Calculate Initial Charges:**
- Charge on \( C_1 \):
\[ Q_1 = C_1 \cdot V_1 = 2 \mu F \cdot 20 V = 40 \mu C \]
- Charge on \( C_2 \):
\[ Q_2 = C_2 \cdot V_2 = 3 \mu F \cdot 40 V = 120 \mu C \]

3. **Connect Capacitors in Series:**
- When connected in series, the positively charged plates are connected together, and the effective charge in the circuit will be determined by the difference in initial charges.
- The initial current \( I_0 \) through the resistance \( R \) can be expressed as:
\[ I_0 = \frac{V_2 - V_1}{R} = \frac{40 V - 20 V}{R} = \frac{20 V}{R} \]

4. **Determine the Charge at Current \( \frac{I_0}{2} \):**
- When the current reduces to \( \frac{I_0}{2} \), we can express the charge on the capacitors:
\[ I = \frac{dQ}{dt} = -\frac{Q_1 + Q_2}{R} \]
- The charge on \( C_1 \) at this instant can be expressed as:
\[ Q_1' = Q_1 + x \]
- The charge on \( C_2 \) at this instant can be expressed as:
\[ Q_2' = Q_2 - x \]

5. **Using Kirchhoff's Voltage Law:**
- The voltage across the capacitors can be expressed as:
\[ V = V_1 + V_2 = \frac{Q_1'}{C_1} + \frac{Q_2'}{C_2} \]
- Substitute \( Q_1' \) and \( Q_2' \):
\[ V = \frac{(Q_1 + x)}{C_1} + \frac{(Q_2 - x)}{C_2} \]

6. **Calculate the Potential Difference Across \( C_1 \):**
- We need to find the potential difference across the 2μF capacitor when the current is \( \frac{I_0}{2} \):
\[ V_{C_1} = \frac{Q_1' + x}{C_1} = \frac{40 \mu C + x}{2 \mu F} \]

7. **Solving for \( x \):**
- At the instant when the current is \( \frac{I_0}{2} \), we can use the relationship derived from the current equations to find \( x \):
\[ x = \frac{(V_2 - V_1)}{2} \cdot \left( \frac{C_1 \cdot C_2}{C_1 + C_2} \right) \]
- Substituting values:
\[ x = \frac{20 V}{2} \cdot \left( \frac{2 \mu F \cdot 3 \mu F}{2 \mu F + 3 \mu F} \right) = 10 V \cdot \left( \frac{6}{5} \right) = 12 \mu C \]

8. **Final Calculation:**
- Now substituting \( x \) back into the equation for \( V_{C_1} \):
\[ V_{C_1} = \frac{40 \mu C + 12 \mu C}{2 \mu F} = \frac{52 \mu C}{2 \mu F} = 26 V \]

### Final Answer:
The potential difference across the 2μF capacitor when the current has reduced to \( \frac{I_0}{2} \) is **26 V**.