• # question_answer An uncharged capacitor of capacitance C is connected with an ideal cell. The emf of the cell is slowly increased from 0 to V (by some mechanism). The total energy taken from the cell in the process of charging of the capacitor is (assume the resistance of the circuit is very small): A) $\frac{1}{2}C{{V}^{2}}$ B) $2C{{V}^{2}}$ C) $\frac{1}{4}C{{V}^{2}}$ D) $C{{V}^{2}}$

The work done by cell$=\int{Vdq}=\int\limits_{0}^{v}{CVdV}=\frac{1}{2}C{{V}^{2}}.$