A circuit consists of two parallel branches connected across an 18 V battery. Branch 1 contains two capacitors C₁ = 2 F and C₂ = 4 F in series. Branch 2 contains two capacitors C₃ = 3 F and C₄ = 6 F in series. The ratio of the electrostatic energy stored in capacitor C₁ to the electrostatic energy stored in capacitor C₄ is:

Step-by-step solution

For Branch 1, capacitors C₁ and C₂ are in series. Their equivalent capacitance is: C_ s1 = C₁ C₂ C₁ + C₂ = 2 4 2 + 4 = 8 6 = 4 3 F . The charge on Branch 1 is: Q₁ = C_ s1 V = 4 3 F 18 V = 24 C . Since C₁ and C₂ are in series, the charge on C₁ is 24 C . The electrostatic energy stored in C₁ is: U₁ = Q₁^2 2C₁ = (24 C )^2 2 2 F = 576 4 J = 144 J . For Branch 2, capacitors C₃ and C₄ are in series. Their equivalent capacitance is: C_ s2 = C₃ C₄ C₃ + C₄ = 3 6 3 + 6 = 18 9 = 2 F . The charge on Branch 2 is: Q₂ = C_ s2 V =