Question

Two capacitors of capacity 4 μF and 6 μF are connected in series to a 500 V battery. The potential difference across the 4 μF capacitor is:

• A. 200 V
• B. 300 V
• C. 400 V
• D. 500 V

Hint:

For series capacitors, use \( \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \) to find the total capacitance, then use the voltage division rule to find the potential difference across each capacitor.

Answer 1

The Correct Option is B

When capacitors are connected in series, the total capacitance \( C_{total} \) is given by: \[ \frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2} \] Substituting the values for \( C_1 = 4 \, \mu F \) and \( C_2 = 6 \, \mu F \): \[ \frac{1}{C_{total}} = \frac{1}{4} + \frac{1}{6} = \frac{5}{12} \quad \Rightarrow \quad C_{total} = \frac{12}{5} = 2.4 \, \mu F \] Now, the total potential difference is distributed across the capacitors. The potential difference across a capacitor in series is given by: \[ V_1 = V_{total} \times \frac{C_{total}}{C_1} \quad \text{and} \quad V_2 = V_{total} \times \frac{C_{total}}{C_2} \] Substitute the known values: \[ V_1 = 500 \times \frac{6}{10} = 300 \, \text{V} \] Thus, the potential difference across the 4 μF capacitor is \( \boxed{300 \, \text{V}} \).

Answer 2

Step 1: Understanding capacitors in series
When capacitors are connected in series, the charge \( Q \) on each capacitor is the same, but the potential difference \( V \) across each capacitor can be different.

Step 2: Formula for capacitors in series
The total voltage \( V_{total} \) is the sum of voltages across each capacitor:
\[ V_{total} = V_1 + V_2 \]
Also, the charge \( Q \) on each capacitor is:
\[ Q = C_1 V_1 = C_2 V_2 \]

Step 3: Given data
Capacitances: \( C_1 = 4 \mu F \), \( C_2 = 6 \mu F \)
Total voltage: \( V_{total} = 500 V \)

Step 4: Calculate voltages across capacitors
Since \( Q = C_1 V_1 = C_2 V_2 \), we have:
\[ V_1 = \frac{Q}{C_1}, \quad V_2 = \frac{Q}{C_2} \]
Adding:
\[ V_1 + V_2 = Q \left(\frac{1}{C_1} + \frac{1}{C_2}\right) = V_{total} \]
Therefore,
\[ Q = \frac{V_{total}}{\frac{1}{C_1} + \frac{1}{C_2}} = \frac{500}{\frac{1}{4} + \frac{1}{6}} = \frac{500}{\frac{5}{12}} = 500 \times \frac{12}{5} = 1200 \mu C \]
Now,
\[ V_1 = \frac{Q}{C_1} = \frac{1200}{4} = 300 V \]

Step 5: Conclusion
The potential difference across the 4 μF capacitor is 300 V.