Five capacitors of capacitances $C_1 = C_2 = C_3 = C_4 = 10 \ \mu\text{F}$ and $C_5 = 2.5 \ \mu\text{F}$ are connected as shown along with a battery of $50 \text{ V}$. 
The equivalent capacitance and the charges on each capacitor respectively are :
Let's analyze the circuit to determine the equivalent capacitance and the charges on each capacitor.
First, observe the arrangement of capacitors:
- Capacitors \( C_1, C_2, C_3, \) and \( C_4 \) are connected in series.
- Series combination is then in parallel with \( C_5 \).
We are given \( C_1 = C_2 = C_3 = C_4 = 10 \ \mu\text{F} \) and \( C_5 = 2.5 \ \mu\text{F} \).
**Step 1: Find the equivalent capacitance of \( C_1, C_2, C_3, \text{and } C_4 \) in series:**
The formula for equivalent capacitance, \( C_{\text{series}} \), of capacitors in series is:
\(C_{\text{series}} = \left( \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \frac{1}{C_4} \right)^{-1}\)
Substituting values:
\(C_{\text{series}} = \left( \frac{1}{10} + \frac{1}{10} + \frac{1}{10} + \frac{1}{10} \right)^{-1} \ \mu\text{F}\)
\(= \left( \frac{4}{10} \right)^{-1} \ \mu\text{F} = \frac{10}{4} \ \mu\text{F} = 2.5 \ \mu\text{F}\)
**Step 2: Find the total equivalent capacitance of the entire circuit:**
The equivalent capacitance of capacitors in parallel is:
\(C_{\text{eq}} = C_{\text{series}} + C_5\)
Substituting the values:
\(C_{\text{eq}} = 2.5 \ \mu\text{F} + 2.5 \ \mu\text{F} = 5 \ \mu\text{F}\)
**Step 3: Calculate the charge on each capacitor:**
Using the formula for charge: \(Q = C \cdot V\)
Total charge for the equivalent capacitor:
\(Q_{\text{total}} = C_{\text{eq}} \cdot V = 5 \ \mu\text{F} \cdot 50 \ \text{V} = 250 \ \mu\text{C}\)
For series capacitors \( C_1, C_2, C_3, \text{and } C_4 \), the charge is the same:
\(Q_1 = Q_2 = Q_3 = Q_4 = Q_{\text{series}} = \frac{Q_{\text{total}}}{2} = 125 \ \mu\text{C}\)
For parallel capacitor \( C_5 \):
\(Q_5 = C_5 \cdot V = 2.5 \ \mu\text{F} \cdot 50 \ \text{V} = 125 \ \mu\text{C}\)
Thus, the answer is that the equivalent capacitance is \(5 \ \mu\text{F}\), and the charge on each capacitor is \(125 \ \mu\text{C}\).
Option: $5 \ \mu\text{F}, 125 \ \mu\text{C}$ on all capacitors.
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