Question

A periodic voltage v (t) = 1 + 4 sin ωt + 2 cos ωt is applied across a 1Ω resistance. The power dissipated is ____________

(a) 1 W

(b) 11 W

(c) 21 W

(d) 24.5 W

Answer 1

Right option is (b) 11 W

For explanation I would say: Given that, v (t) = 1 + 4 sin ωt + 2 cos ωt

So, Power is given by,

Power, P = \(\frac{1^2}{1} + \frac{\frac{4^2}{\sqrt{2}}}{1} +  \frac{\frac{2^2}{\sqrt{2}}}{1}\)

= 11 W.