A capacitor of capacitance `C_(1)=1 muF` withstand a maximum voltage of `V_(1)=6 KV`, and another capacitor of capacitance `C_(2)=2 muF`, can with stand a maximum voltage of `V_(2)=4KV`. If they are connected in series, what maximum voltage will the system withstand?

Explanation

To solve the problem step by step, we will 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 \( C_2 = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- Maximum Voltage \( V_2 = 4 \, kV = 4000 \, V \)

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

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

For Capacitor 2:
\[ Q_2 = C_2 \times V_2 = (2 \times 10^{-6} \, F) \times (4000 \, V) = 8 \times 10^{-3} \, C \]

### Step 3: Determine the maximum charge the series combination can withstand
When capacitors are connected in series, the charge \( Q \) on each capacitor is the same. Therefore, the maximum charge that the series combination can withstand is determined by the capacitor with the lower maximum charge.

From Step 2:
- \( Q_1 = 6 \times 10^{-3} \, C \)
- \( Q_2 = 8 \times 10^{-3} \, C \)

The maximum charge \( Q_{max} \) for the series combination is:
\[ Q_{max} = \min(Q_1, Q_2) = 6 \times 10^{-3} \, C \]

### Step 4: Calculate the total voltage across the series combination
The total voltage \( V \) across the series combination of capacitors can be calculated using the formula:
\[ V = \frac{Q_{max}}{C_{eq}} \]
where \( C_{eq} \) is the equivalent capacitance of the capacitors in series.

The equivalent capacitance \( C_{eq} \) for capacitors in series is given by:
\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \]

Calculating \( C_{eq} \):
\[ \frac{1}{C_{eq}} = \frac{1}{1 \times 10^{-6}} + \frac{1}{2 \times 10^{-6}} \]
\[ \frac{1}{C_{eq}} = 1000000 + 500000 = 1500000 \]
\[ C_{eq} = \frac{1}{1500000} \approx 6.67 \times 10^{-7} \, F \]

Now substituting \( Q_{max} \) and \( C_{eq} \) into the voltage formula:
\[ V = \frac{6 \times 10^{-3} \, C}{6.67 \times 10^{-7} \, F} \]
\[ V \approx 9000 \, V = 9 \, kV \]

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

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