Question:

Effective capacitance of parallel combination of two capacitors \(C_1\) and \(C_2\) is 10 \(\mu\)F. When these capacitors are individually connected to a voltage source of 1V, the energy stored in the capacitor \(C_2\) is 4 times that of \(C_1\). If these capacitors are connected in series, their effective capacitance will be :

Show Hint

Energy stored at fixed voltage is proportional to capacitance, so C2 = 4 C1; combine with C1 + C2 = 10 uF.
Updated On: Nov 5, 2026
  • 4.2 \(\mu\)F
  • 8.4 \(\mu\)F
  • 1.6 \(\mu\)F
  • 3.2 \(\mu\)F
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Parallel: \(C_1 + C_2 = 10\ \mu F\).
Energy \(U = \frac{1}{2}CV^2\) with the same V = 1 V, so \(U_2 = 4U_1 \Rightarrow C_2 = 4C_1\).
\(5C_1 = 10 \Rightarrow C_1 = 2\ \mu F\), \(C_2 = 8\ \mu F\).
Series: \(C_s = \frac{C_1C_2}{C_1+C_2} = \frac{2 \times 8}{10} = 1.6\ \mu F\).
Answer: (3) 1.6 \(\mu\)F
Was this answer helpful?
0
0

Top JEE Main Physics Questions

View More Questions

Top JEE Main Capacitors and Capacitance Questions

View More Questions