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Two identical capacitors C1 and C2 of equal capacitance are connected as shown in the circuit. Terminals a and b of the key k are connected to charge capacitor C1 using battery of emf V volt. Now disconnecting a and b the terminals b and c are connected. Due to this, what will be the percentage loss of energy?
Detailed Solution
$U_i=\frac{1}{2}CV^2$. Loss on connecting $=\frac{C\cdot C}{2(C+C)}(V-0)^2=\frac{1}{4}CV^2$. Percentage loss $=\frac{CV^2/4}{CV^2/2}\times100=50\%$.