An unchanged capacitor is connected to a battery. How much amount of energy supplied by the battery is lost as heat while charging the capacitor?

$\frac{1}{2}QV$

The correct answer is Option (3) → $\frac{1}{2}QV$

$E_{Battery}=QV$

where,

$Q=CV$ [Charge on capacitor]

V = Voltage of the battery

C = Capacitance of capacitor

$∴E_{Battery}=CV^2$

Energy stored in the capacitor when its fully charged

$E_{capacitor}=\frac{1}{2}CV^2$ [formula]

$∴E_{heat}=CV^2-\frac{1}{2}CV^2=\frac{1}{2}CV^2=\frac{1}{2}QV$