A capacitor of capacitance C and potential V has energy E. It is connected to another capacitor of capacitance 2 C and potential 2V. Then the loss of energy is (x/3)E, where x i s _____ .

Question

G1 Physics · Accepted Answer

Energy loss when two charged capacitors are connected in parallel: ΔE = ½ [C₁C₂/(C₁ + C₂)] (V₁ - V₂)² Here, C₁ = C C₂ = 2C V₁ = V V₂ = 2V Therefore, ΔE = ½ [2C²/(3C)]V² = ½(2C/3)V² = CV²/3 Given initial energy of first capacitor: E = ½CV² So, ΔE = (2/3)E Comparing with: ΔE = (x/3)E x = 2