Three capacitors with capacitances of `1muF, 2muF and 3muF` are connected in series. Each capacitor gets punctured, if a potential difference just exceeding 100 volt is applied. If the group is connected across 220 volt circuit then the capacitor most likely to puncture first is

Options

A. capacitance `1muF`
B. capacitance `2muF`
C. capacitance `3muF`
D. capacitance `1muF` (or) `2muF` (or) `3muF`

Answer

To solve the problem, we need to determine which capacitor is most likely to puncture first when connected in series across a 220V circuit. The capacitors have capacitances of 1μF, 2μF, and 3μF.

### Step-by-Step Solution:

1. **Understand Capacitors in Series**:
When capacitors are connected in series, the total capacitance (C_total) can be calculated using the formula:
\[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \]
where \(C_1 = 1μF\), \(C_2 = 2μF\), and \(C_3 = 3μF\).

2. **Calculate Total Capacitance**:
Plugging in the values:
\[ \frac{1}{C_{total}} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} \]
Finding a common denominator (6):
\[ \frac{1}{C_{total}} = \frac{6}{6} + \frac{3}{6} + \frac{2}{6} = \frac{11}{6} \]
Therefore,
\[ C_{total} = \frac{6}{11} μF \approx 0.545 μF \]

3. **Calculate Charge on Each Capacitor**:
The charge (Q) stored in a capacitor is given by:
\[ Q = C \cdot V \]
In a series connection, the charge on each capacitor is the same. The voltage across each capacitor can be found using:
\[ V_i = \frac{Q}{C_i} \]
where \(V_i\) is the voltage across capacitor \(i\).

4. **Calculate the Voltage Across Each Capacitor**:
First, we need to find the total charge when the circuit is connected to 220V:
\[ Q = C_{total} \cdot V = \left(\frac{6}{11} μF\right) \cdot 220V = \frac{1320}{11} μC = 120 μC \]
Now, calculate the voltage across each capacitor:
- For \(C_1 = 1μF\):
\[ V_1 = \frac{Q}{C_1} = \frac{120 μC}{1 μF} = 120 V \]
- For \(C_2 = 2μF\):
\[ V_2 = \frac{Q}{C_2} = \frac{120 μC}{2 μF} = 60 V \]
- For \(C_3 = 3μF\):
\[ V_3 = \frac{Q}{C_3} = \frac{120 μC}{3 μF} = 40 V \]

5. **Determine Which Capacitor Punctures First**:
Each capacitor can withstand a maximum voltage of just over 100V before puncturing. From our calculations:
- \(V_1 = 120 V\) (exceeds 100V)
- \(V_2 = 60 V\) (does not exceed 100V)
- \(V_3 = 40 V\) (does not exceed 100V)

Since the first capacitor (1μF) experiences a voltage of 120V, which exceeds its puncture threshold of 100V, it is the most likely to puncture first.

### Conclusion:
The capacitor most likely to puncture first is the **1μF capacitor**.