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 step by step, we will follow the process of analyzing the circuit with the two capacitors and applying Kirchhoff's laws.

### Step 1: Calculate the initial charge on each capacitor.
The charge \( Q \) on a capacitor is given by the formula:
\[ Q = C \times V \]
For the \( 2 \mu F \) capacitor charged to \( 20 V \):
\[ Q_1 = 2 \mu F \times 20 V = 40 \mu C \]
For the \( 3 \mu F \) capacitor charged to \( 40 V \):
\[ Q_2 = 3 \mu F \times 40 V = 120 \mu C \]

### Step 2: Analyze the circuit when the capacitors are connected.
When the capacitors are connected in series with their positively charged plates connected together, the total charge \( Q \) in the circuit will be the difference of the two charges:
\[ Q = Q_2 - Q_1 = 120 \mu C - 40 \mu C = 80 \mu C \]
This charge will redistribute between the two capacitors.

### Step 3: Apply Kirchhoff's Voltage Law (KVL) for the initial state.
Initially, the current through the resistor is \( I_0 \). We can write the KVL equation for the loop:
\[ I_0 R - \frac{Q_1}{2 \mu F} - \frac{Q_2}{3 \mu F} = 0 \]
Substituting \( Q_1 = 40 \mu C \) and \( Q_2 = 120 \mu C \):
\[ I_0 R - \frac{40}{2} - \frac{120}{3} = 0 \]
This simplifies to:
\[ I_0 R - 20 - 40 = 0 \implies I_0 R = 60 \]

### Step 4: Analyze the circuit when the current has reduced to \( \frac{I_0}{2} \).
At this point, let \( Q \) be the charge on the \( 2 \mu F \) capacitor. The KVL equation becomes:
\[ \frac{I_0}{2} R - \frac{40 + Q}{2 \mu F} - \frac{120 - Q}{3 \mu F} = 0 \]
Substituting \( I_0 R = 60 \):
\[ 30 - \frac{40 + Q}{2} - \frac{120 - Q}{3} = 0 \]

### Step 5: Solve for \( Q \).
Multiplying through by 6 to eliminate the denominators:
\[ 180 - 3(40 + Q) - 2(120 - Q) = 0 \]
Expanding gives:
\[ 180 - 120 - 3Q - 240 + 2Q = 0 \]
Combining like terms:
\[ -3Q + 2Q = 180 - 120 - 240 \]
This simplifies to:
\[ -Q = -180 \implies Q = 60 \mu C \]

### Step 6: Calculate the potential difference across the \( 2 \mu F \) capacitor.
The potential difference \( V \) across the \( 2 \mu F \) capacitor is given by:
\[ V = \frac{Q + 40 \mu C}{2 \mu F} = \frac{60 + 40}{2} = \frac{100}{2} = 50 V \]

### Final Answer
The potential difference across the \( 2 \mu F \) capacitor at the instant the current has reduced to \( \frac{I_0}{2} \) is **50 V**.