A capacitor of capacitance `C_1 = 1.0muF` charged upto a voltage `V = 110 V` is connected in parallel to the terminals of a circuit consisting of two uncharged capacitors connected in series and possessing the capacitance `C_2 = 2.0 muF` and `C_3 = 3.0muF`. What charge will flow through the connecting wires?

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the initial charge on capacitor \( C_1 \)

The charge \( Q_1 \) on capacitor \( C_1 \) can be calculated using the formula:
\[ Q_1 = C_1 \times V \]
Where:
- \( C_1 = 1.0 \, \mu F = 1.0 \times 10^{-6} \, F \)
- \( V = 110 \, V \)

Substituting the values:
\[ Q_1 = 1.0 \times 10^{-6} \, F \times 110 \, V = 1.1 \times 10^{-4} \, C = 110 \, \mu C \]

### Step 2: Find the equivalent capacitance of capacitors \( C_2 \) and \( C_3 \) in series

The equivalent capacitance \( C_{eq} \) for capacitors in series is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_2} + \frac{1}{C_3} \]
Where:
- \( C_2 = 2.0 \, \mu F = 2.0 \times 10^{-6} \, F \)
- \( C_3 = 3.0 \, \mu F = 3.0 \times 10^{-6} \, F \)

Calculating \( C_{eq} \):
\[ \frac{1}{C_{eq}} = \frac{1}{2.0 \times 10^{-6}} + \frac{1}{3.0 \times 10^{-6}} = \frac{3 + 2}{6} \times 10^{6} = \frac{5}{6} \times 10^{6} \]
Thus,
\[ C_{eq} = \frac{6}{5} \times 10^{-6} \, F = 1.2 \, \mu F \]

### Step 3: Apply charge conservation in the parallel circuit

When \( C_1 \) is connected in parallel with the equivalent capacitance \( C_{eq} \), the total charge \( Q_{total} \) is conserved:
\[ Q_1 = Q_{eq} + Q_{C1} \]
Where \( Q_{C1} \) is the charge on \( C_1 \) after connecting to the uncharged capacitors.

### Step 4: Relate the charges using the capacitance values

Since both capacitors are in parallel, the voltage across them will be the same:
\[ V = \frac{Q_{C1}}{C_1} = \frac{Q_{eq}}{C_{eq}} \]

From this, we can express \( Q_{C1} \) in terms of \( Q_{eq} \):
\[ Q_{C1} = C_1 \times V \quad \text{and} \quad Q_{eq} = C_{eq} \times V \]

### Step 5: Set up the equations

From the conservation of charge:
\[ Q_1 = Q_{C1} + Q_{eq} \]
Substituting \( Q_{C1} \) and \( Q_{eq} \):
\[ 110 \, \mu C = \frac{C_1}{C_{eq}} Q_{eq} + Q_{eq} \]
This simplifies to:
\[ 110 \, \mu C = \left( \frac{C_1 + C_{eq}}{C_{eq}} \right) Q_{eq} \]

### Step 6: Solve for \( Q_{eq} \)

Substituting \( C_1 = 1.0 \, \mu F \) and \( C_{eq} = 1.2 \, \mu F \):
\[ 110 \, \mu C = \left( \frac{1.0 + 1.2}{1.2} \right) Q_{eq} \]
\[ 110 \, \mu C = \left( \frac{2.2}{1.2} \right) Q_{eq} \]
\[ Q_{eq} = \frac{110 \times 1.2}{2.2} \, \mu C = 60 \, \mu C \]

### Final Answer

The charge that flows through the connecting wires is \( Q_{eq} = 60 \, \mu C \).

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