A capacitor of capacitance `1 muF` withstands the maximum voltage 6 kV while a capacitor of `2muF` withstands the maximum voltage 4 kV. What maximum voltage will the system of these two capacitor withstands if they are connected in series?

Options

A. 10 kV
B. 12 kV
C. 8 kV
D. 9 kV

Answer

To solve the problem of finding the maximum voltage that a system of two capacitors can withstand when connected in series, we can follow these steps:

### Step 1: Identify the parameters of the capacitors
We have two capacitors:
- Capacitor 1 (C1): Capacitance = 1 µF, Maximum Voltage (V1) = 6 kV
- Capacitor 2 (C2): Capacitance = 2 µF, Maximum Voltage (V2) = 4 kV

### Step 2: Calculate the charge on each capacitor
The charge (Q) on a capacitor is given by the formula:
\[ Q = C \times V \]

For Capacitor 1:
\[ Q1 = C1 \times V1 = 1 \times 10^{-6} \, F \times 6 \times 10^{3} \, V = 6 \times 10^{-3} \, C \]

For Capacitor 2:
\[ Q2 = C2 \times V2 = 2 \times 10^{-6} \, F \times 4 \times 10^{3} \, V = 8 \times 10^{-3} \, C \]

### Step 3: Determine the charge in series connection
When capacitors are connected in series, the charge on each capacitor is the same. Therefore, the charge (Q) in the series connection will be limited by the capacitor with the lower maximum charge capacity. In this case, the charge on the first capacitor (Q1) is less than the charge on the second capacitor (Q2).

Thus, we take:
\[ Q_{series} = Q1 = 6 \times 10^{-3} \, C \]

### Step 4: Calculate the equivalent capacitance of the series connection
The equivalent capacitance (C_eq) of capacitors in series is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C1} + \frac{1}{C2} \]

Calculating this:
\[ \frac{1}{C_{eq}} = \frac{1}{1 \times 10^{-6}} + \frac{1}{2 \times 10^{-6}} \]
\[ \frac{1}{C_{eq}} = 1 \times 10^{6} + 0.5 \times 10^{6} = 1.5 \times 10^{6} \]
\[ C_{eq} = \frac{1}{1.5 \times 10^{6}} = \frac{2}{3} \times 10^{-6} \, F \]

### Step 5: Calculate the maximum voltage across the series combination
The maximum voltage (V_max) across the series combination can be calculated using:
\[ V_{max} = \frac{Q_{series}}{C_{eq}} \]

Substituting the values:
\[ V_{max} = \frac{6 \times 10^{-3} \, C}{\frac{2}{3} \times 10^{-6} \, F} \]
\[ V_{max} = 6 \times 10^{-3} \times \frac{3}{2} \times 10^{6} \]
\[ V_{max} = 9 \times 10^{3} \, V = 9 \, kV \]

### Final Answer
The maximum voltage that the system of these two capacitors can withstand when connected in series is **9 kV**.

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