Three capacitors `2muF, 3muF` and `5muF` can withstand voltages to `3V,2V` and `1V` respectively. Their series combination can withstand a maximum voltage equal to

Options

A. `5` Volts
B. `(31//6)` Volts
C. `(26//5)` Volts
D. None

Explanation

To find the maximum voltage that the series combination of the three capacitors can withstand, we need to analyze the individual capacitors and their voltage ratings.

### Step-by-Step Solution:

1. **Identify the Capacitors and Their Ratings**:
- Capacitor C1: \(2 \mu F\) with a maximum voltage of \(3 V\)
- Capacitor C2: \(3 \mu F\) with a maximum voltage of \(2 V\)
- Capacitor C3: \(5 \mu F\) with a maximum voltage of \(1 V\)

2. **Determine the Charge on Each Capacitor**:
- The charge \(Q\) on each capacitor in series is the same. The maximum charge that each capacitor can hold can be calculated using the formula:
\[ Q = C \times V \]
- For C1:
\[ Q_1 = C_1 \times V_{max1} = 2 \mu F \times 3 V = 6 \mu C \]
- For C2:
\[ Q_2 = C_2 \times V_{max2} = 3 \mu F \times 2 V = 6 \mu C \]
- For C3:
\[ Q_3 = C_3 \times V_{max3} = 5 \mu F \times 1 V = 5 \mu C \]

3. **Identify the Limiting Factor**:
- In a series circuit, the charge \(Q\) must be the same across all capacitors. The capacitor with the smallest charge capacity limits the total charge in the circuit.
- Here, C3 can only handle \(5 \mu C\), which is the smallest charge capacity.

4. **Calculate the Maximum Voltage Across Each Capacitor**:
- Now we can calculate the voltage across each capacitor when \(Q = 5 \mu C\):
- For C1:
\[ V_1 = \frac{Q}{C_1} = \frac{5 \mu C}{2 \mu F} = 2.5 V \]
- For C2:
\[ V_2 = \frac{Q}{C_2} = \frac{5 \mu C}{3 \mu F} \approx 1.67 V \]
- For C3:
\[ V_3 = \frac{Q}{C_3} = \frac{5 \mu C}{5 \mu F} = 1 V \]

5. **Calculate the Total Voltage in Series**:
- The total voltage \(V_{total}\) across the series combination is the sum of the voltages across each capacitor:
\[ V_{total} = V_1 + V_2 + V_3 = 2.5 V + 1.67 V + 1 V \approx 5.17 V \]

6. **Conclusion**:
- The maximum voltage that the series combination of the capacitors can withstand is approximately \(5.17 V\).