Question

Two capacitors of capacitances \( 4 \, \mu\text{F} \) and \( 6 \, \mu\text{F} \) are connected in series across a potential difference of \( 300 \, \text{V} \). What is the potential difference across the \( 4 \, \mu\text{F} \) capacitor?

• A. \( 100 \, \text{V} \)
• B. \( 120 \, \text{V} \)
• C. \( 180 \, \text{V} \)
• D. \( 200 \, \text{V}

Hint:

\textbf{Key Fact:} In series, the voltage divides inversely with capacitance: smaller capacitance gets a larger voltage share.

Answer 1

The Correct Option is C

• Equivalent Capacitance in Series: For capacitors in series, the equivalent capacitance C_eq is given by 1/C_eq = 1/C₁ + 1/C₂. Here, C₁ = 4 μF, C₂ = 6 μF.
  1/C_eq = 1/4 + 1/6 = (3 + 2)/12 = 5/12, so C_eq = 12/5 = 2.4 μF.
• Charge Stored: The total charge Q on the equivalent capacitor is Q = C_eq × V = 2.4 × 300 = 720 μC.
• Voltage Across Each Capacitor: In series, the charge is the same across both capacitors. Voltage across a capacitor is V = Q/C.
  • For 4 μF: V₁ = Q/C₁ = 720/4 = 180 V.
  • For 6 μF: V₂ = Q/C₂ = 720/6 = 120 V.
• Verify Total Voltage: V₁ + V₂ = 180 + 120 = 300 V, which matches the applied voltage.
• Conclusion: The potential difference across the 4 μF capacitor is (3) 180 V.