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 V
B. 31/6 V
C. 26/5 V
D. 6 V

Answer

To solve the problem of finding the maximum voltage that can be withstood by the series combination of three capacitors (2 µF, 3 µF, and 5 µF), we will follow these steps:

### Step 1: Determine the maximum charge for each capacitor
We start by calculating the maximum charge that each capacitor can store based on their capacitance and the maximum voltage they can withstand.

1. **For the 2 µF capacitor:**
\[ Q_1 = C_1 \times V_1 = 2 \, \mu F \times 3 \, V = 6 \, \mu C \]

2. **For the 3 µF capacitor:**
\[ Q_2 = C_2 \times V_2 = 3 \, \mu F \times 2 \, V = 6 \, \mu C \]

3. **For the 5 µF capacitor:**
\[ Q_3 = C_3 \times V_3 = 5 \, \mu F \times 1 \, V = 5 \, \mu C \]

### Step 2: Identify the limiting charge
In a series combination of capacitors, the charge stored on each capacitor is the same. Therefore, the maximum charge that can be stored in the series combination is determined by the capacitor with the smallest maximum charge.

The maximum charges calculated are:
- \( Q_1 = 6 \, \mu C \)
- \( Q_2 = 6 \, \mu C \)
- \( Q_3 = 5 \, \mu C \)

Thus, the limiting charge \( Q_{max} \) is:
\[ Q_{max} = \min(Q_1, Q_2, Q_3) = 5 \, \mu C \]

### Step 3: Calculate the equivalent capacitance
Next, we calculate the equivalent capacitance \( C_{eq} \) of the series combination using the formula:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
Substituting the values:
\[ \frac{1}{C_{eq}} = \frac{1}{2 \, \mu F} + \frac{1}{3 \, \mu F} + \frac{1}{5 \, \mu F} \]
Finding a common denominator (which is 30):
\[ \frac{1}{C_{eq}} = \frac{15}{30} + \frac{10}{30} + \frac{6}{30} = \frac{31}{30} \]
Thus,
\[ C_{eq} = \frac{30}{31} \, \mu F \]

### Step 4: Calculate the maximum voltage across the series combination
Using the relationship between charge, capacitance, and voltage:
\[ Q = C_{eq} \times V_{max} \]
We can rearrange this to find \( V_{max} \):
\[ V_{max} = \frac{Q_{max}}{C_{eq}} \]
Substituting the known values:
\[ V_{max} = \frac{5 \, \mu C}{\frac{30}{31} \, \mu F} = 5 \times \frac{31}{30} \, V = \frac{155}{30} \, V = \frac{31}{6} \, V \]

### Conclusion
The maximum voltage that can be withstood by the series combination of the three capacitors is:
\[ V_{max} = \frac{31}{6} \, V \approx 5.17 \, V \]