A capacitor C₁ of 4 F is connected in series with an unknown capacitor C₂ across a DC voltage source. The energy stored in C₁ is found to be 4 times the energy stored in C₂ . The capacitance of C₂ is :

Step-by-step solution

In a series combination, the charge Q on both capacitors is the same. The energy stored in a capacitor is given by U = Q^2 2C . This shows that for a constant charge, the energy stored is inversely proportional to the capacitance ( U 1 C ). Given that the energy stored in C₁ is 4 times the energy stored in C₂ ( U₁ = 4 U₂ ), we can write: Q^2 2C₁ = 4 ( Q^2 2C₂ ) Simplifying this, we get: 1 C₁ = 4 C₂ Rearranging for C₂ gives: C₂ = 4 C₁ Substituting the given value C₁ = 4 F : C₂ = 4 4 F = 16 F Answer: 16 F