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 phi = (E_(0))/(sqrt2) xx (I_(0))/(sqrt2) xx (R)/(Z)`
`Rightarrow (E_(0))/(sqrt2) xx (E_(0))/(Zsqrt2) xx (R)/(Z) rightarrow P = (E_(0)^(2)R)/(2Z^(2))`
Given `X_(L) = R` so `Z = sqrt2 R rightarrow P = (E_(0)^(2)R)/(4R^(2)) = (E_(0)^(2))/(4R)`