A battery of emf 20V is connect to two capacitors `1 mu F` and `3 mu F` in series. `1 mu F` capacitor withstands a maximum of 9V and `3 mu F` withstands a maximum voltage of 6V then

Options

A. `1 mu F ` capacitors break
B. `2 mu F ` capacitors break
C. both will break
D. no capacitor will break

Answer

To solve the problem step by step, we will analyze the given information and calculate the voltages across the capacitors connected in series.

### Step 1: Understand the Configuration
We have two capacitors connected in series:
- Capacitor 1 (C1) = 1 µF
- Capacitor 2 (C2) = 3 µF
- Battery emf (V) = 20 V

### Step 2: Calculate the Equivalent Capacitance
For capacitors in series, the equivalent capacitance (C_eq) can be calculated using the formula:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]
Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{1 \mu F} + \frac{1}{3 \mu F} = \frac{3 + 1}{3 \mu F} = \frac{4}{3 \mu F} \]
Thus,
\[ C_{eq} = \frac{3}{4} \mu F \]

### Step 3: Calculate the Charge on the Capacitors
The charge (Q) stored in the equivalent capacitor when connected to the battery can be calculated using:
\[ Q = C_{eq} \times V \]
Substituting the values:
\[ Q = \left(\frac{3}{4} \times 10^{-6} F\right) \times 20 V = 15 \times 10^{-6} C = 15 \mu C \]

### Step 4: Calculate the Voltage Across Each Capacitor
1. **Voltage across Capacitor 1 (V1)**:
\[ V_1 = \frac{Q}{C_1} = \frac{15 \mu C}{1 \mu F} = 15 V \]

2. **Voltage across Capacitor 2 (V2)**:
\[ V_2 = \frac{Q}{C_2} = \frac{15 \mu C}{3 \mu F} = 5 V \]

### Step 5: Check the Maximum Voltage Ratings
- Capacitor 1 can withstand a maximum of 9 V, but it has 15 V across it. Therefore, it will **break down**.
- Capacitor 2 can withstand a maximum of 6 V, but it has only 5 V across it. Therefore, it will **not break down**.

### Conclusion
The 1 µF capacitor will break down because it exceeds its maximum voltage rating, while the 3 µF capacitor will function properly.

### Final Answer
**The 1 µF capacitor will break down.**

---