A capacitor 1mF withstands a maximum voltage of 6KV while another capacitor 2mF withstands a maximum voltage of 4KV. If the capacitors are connected in series, the system will withstand a maximum voltage of

Options

A. `2KV `
B. `4KV`
C. `6KV`
D. `9KV`

Answer

To solve the problem, we need to find the maximum voltage that the system of capacitors can withstand when they are connected in series. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Capacitor 1 (C1): 1 mF (1 millifarad) with a maximum voltage (V1) of 6 kV (6000 volts).
- Capacitor 2 (C2): 2 mF (2 millifarads) with a maximum voltage (V2) of 4 kV (4000 volts).

### 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 (C1):
\[ Q1 = C1 \times V1 = (1 \times 10^{-3} \, \text{F}) \times (6 \times 10^{3} \, \text{V}) \]
\[ Q1 = 6 \, \text{C} \]

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

### Step 3: Determine the maximum charge in series
When capacitors are connected in series, the charge on each capacitor is the same. The maximum charge that the series combination can withstand is determined by the capacitor with the lower charge capacity. Here, since \( Q1 < Q2 \):
\[ Q_{\text{max}} = Q1 = 6 \, \text{C} \]

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

Substituting the values:
\[ \frac{1}{C_{\text{eq}}} = \frac{1}{1 \times 10^{-3}} + \frac{1}{2 \times 10^{-3}} \]
\[ \frac{1}{C_{\text{eq}}} = 1000 + 500 = 1500 \]
\[ C_{\text{eq}} = \frac{1}{1500} = \frac{2}{3} \times 10^{-3} \, \text{F} \]

### Step 5: Calculate the maximum voltage across the equivalent capacitance
The maximum voltage (V_max) that the series combination can withstand is given by:
\[ V_{\text{max}} = \frac{Q_{\text{max}}}{C_{\text{eq}}} \]

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

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

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