Correct Option: A

Since energy will be maximum when
sin 2πt = 1

sin 2πt = sin	π
	2

t =	1	sec
	4

Power dissipated
= I² R = (10^{– 4} sin 2πt)². 10⁶ = 10^{– 2} sin² 2πt
Power dissipated

=		1 / 2	(10–2 sin2 2πt) dt
		0

=		1 / 2	1	(1 – cos 4πt) dt
		0	2

= 2.5 × 10– 3 joule
Note:

Since the current flows in resistor for		1	sec during charging and	1	sec during discharging so
		4		4

t =		1	+	1	=	1	sec.
		4		4		2

∴ Power dissipated from 0 to	1	sec .
	2