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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 isA. `(E_(0)^(2))/(R)`B. `(E_(0)^(2))/(2R)`C. `(E_(0)^(2))/(4R)`D. `(E_(0)^(2))/(8R)` |
Answer» Correct Answer - C `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)` |
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