In an `LR`-circuit, the inductive reactance is equal to the resistance `R` of the circuit. An e.m.f `E=E_(0)cos (omegat)` applied to the circuit. The power consumed in the circuit is

Options

A. `(E_(0)^(2))/(R)`
B. `(E_(0)^(2))/(2R)`
C. `(E_(0)^(2))/(4R)`
D. `(E_(0)^(2))/(8R)`

Answer

C

Explanation

`P=E_("rms")i_("rms")cos varphi=(E_(0))/(sqrt(2))xx(i_(0))/(sqrt(2))xxR/Z`
`implies (E_(0))/(sqrt(2))xx(E_(0))/(Zsqrt(2))xxR/Z implies P=(E_(0)^(2)R)/(2Z^(2))`
Given `X_(L)=R`, so `Z=sqrt(2)R implies P=(E_(0)^(2))/(4R)`