A capacitor of capacitance `C_(1)=1 muF` can withstand maximum voltage `V_(1)=6 kV` and another capacitor of capacitance `C_(2)=3 muF` can withstand maximum voltage `V_(2) 4 kV`. When the two capacitors are connected in series, the combined system can withstand a maximum voltage of

Options

A. `4kV`
B. `6kV`
C. `8kV`
D. `10kV`

Explanation

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

### Step 1: Identify the given values
We have two capacitors:
- Capacitor 1:
- Capacitance \( C_1 = 1 \, \mu F = 1 \times 10^{-6} \, F \)
- Maximum Voltage \( V_1 = 6 \, kV = 6000 \, V \)

- Capacitor 2:
- Capacitance \( C_2 = 3 \, \mu F = 3 \times 10^{-6} \, F \)
- Maximum Voltage \( V_2 = 4 \, kV = 4000 \, V \)

### Step 2: Calculate the maximum charge for each capacitor
The maximum charge \( Q \) that can be stored in a capacitor is given by the formula:
\[ Q = C \times V \]

For Capacitor 1:
\[ Q_1 = C_1 \times V_1 = (1 \times 10^{-6} \, F) \times (6000 \, V) = 6 \times 10^{-3} \, C = 6 \, mC \]

For Capacitor 2:
\[ Q_2 = C_2 \times V_2 = (3 \times 10^{-6} \, F) \times (4000 \, V) = 12 \times 10^{-3} \, C = 12 \, mC \]

### Step 3: Determine the limiting charge when capacitors are in series
When capacitors are connected in series, the charge \( Q \) on each capacitor is the same. Therefore, the maximum charge that can be stored in the series combination is limited by the capacitor with the smaller maximum charge.

In this case, \( Q = Q_1 = 6 \, mC \) (since \( Q_1 < Q_2 \)).

### Step 4: Calculate the voltage across each capacitor with the common charge
Now we can calculate the voltage across each capacitor when the charge is \( Q = 6 \, mC \).

For Capacitor 1:
\[ V_{1}' = \frac{Q}{C_1} = \frac{6 \times 10^{-3} \, C}{1 \times 10^{-6} \, F} = 6000 \, V = 6 \, kV \]

For Capacitor 2:
\[ V_{2}' = \frac{Q}{C_2} = \frac{6 \times 10^{-3} \, C}{3 \times 10^{-6} \, F} = 2000 \, V = 2 \, kV \]

### Step 5: Calculate the total voltage across the series combination
The total voltage \( V_{total} \) that the series combination can withstand is the sum of the voltages across each capacitor:
\[ V_{total} = V_{1}' + V_{2}' = 6000 \, V + 2000 \, V = 8000 \, V = 8 \, kV \]

### Final Answer:
The combined system can withstand a maximum voltage of **8 kV**.

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