Correct Option: A

Using addition of admittances in parallel,

Y = YL + YR + YC =	1	+ jωC = 1 + j
	jωL

The phasor voltage becomes

V =	1
	Y

Using phasors in polar format,
I = 10 e^{– jπ / 2}
Y = √2 e^{– jπ / 4}
Power (time average power) dissipated in R,

P =	1	Re	(VV*)	= 25
	2		R

υ(t) =	10	cos		ωt -	3π
	√2				4

from which, P = (v)² (t) = 25 watts.