Two capacitor having capacitances `8muF` and `16 muF` have breaking voltage `20V` and `80V`. They are combined in series. The maximum charge they can store individually in the combination is

Options

A. `160 muC`
B. `200 muC`
C. `320 muC`
D. `480 muC`

Answer

To solve the problem of finding the maximum charge that two capacitors can store when combined in series, we will follow these steps:

### Step 1: Identify the given values
- Capacitance of the first capacitor, \( C_1 = 8 \mu F \)
- Breaking voltage of the first capacitor, \( V_1 = 20 V \)
- Capacitance of the second capacitor, \( C_2 = 16 \mu F \)
- Breaking voltage of the second capacitor, \( V_2 = 80 V \)

### Step 2: Calculate the maximum charge for each capacitor
The maximum charge \( Q \) that a capacitor can store is given by the formula:
\[ Q = C \times V \]
where \( C \) is the capacitance and \( V \) is the voltage.

#### For the first capacitor:
\[ Q_1 = C_1 \times V_1 = 8 \mu F \times 20 V \]
\[ Q_1 = 8 \times 10^{-6} F \times 20 V = 160 \mu C \]

#### For the second capacitor:
\[ Q_2 = C_2 \times V_2 = 16 \mu F \times 80 V \]
\[ Q_2 = 16 \times 10^{-6} F \times 80 V = 1280 \mu C \]

### Step 3: Determine the maximum charge in series combination
In a series combination of capacitors, the charge \( Q \) on each capacitor is the same. Therefore, the maximum charge that can be stored in the combination is limited by the capacitor with the smaller charge capacity.

Since \( Q_1 = 160 \mu C \) is less than \( Q_2 = 1280 \mu C \), the maximum charge that can be stored in the series combination is:
\[ Q_{max} = Q_1 = 160 \mu C \]

### Conclusion
The maximum charge that the capacitors can store individually in the combination is \( 160 \mu C \).