Question

in a series rlc circuit with resistance r and impedance z, the rms value of the generator emf is
( varepsilon _ { mathrm { rms } } ) and the rms value of the current is ( i _ { mathrm { rms } } ). the current lags the emf by ( phi ). the average power supplied by the generator is given by
( \begin{array} { l l l l l } { \bigcirc i _ { mathrm { rms } } ^ { 2 } r } & { \bigcirc i _ { mathrm { rms } } varepsilon _ { mathrm { rms } } } & { \bigcirc i _ { mathrm { rms } } ^ { 2 } z } & { \bigcirc i _ { mathrm { rms } } ^ { 2 } / z } & { \bigcirc \frac { i _ { mathrm { rms } } varepsilon _ { mathrm { rms } } } { 2 cos phi } } end{array} )

Explanation:

Step1: Recall the formula for average power in AC circuits

The average power \( P_{avg} \) in an AC circuit is given by \( P_{avg}=i_{rms}\varepsilon_{rms}\cos\phi \). Also, from Ohm's law for AC circuits \( \varepsilon_{rms}=i_{rms}Z \).

Step2: Substitute \( \varepsilon_{rms}=i_{rms}Z \) into the power formula

Substituting \( \varepsilon_{rms}=i_{rms}Z \) into \( P_{avg}=i_{rms}\varepsilon_{rms}\cos\phi \), we get \( P_{avg}=i_{rms}^2Z\cos\phi \). But another way to express power is \( P_{avg}=i_{rms}^2R \) (since the average power is dissipated only in the resistor, and \( R = Z\cos\phi \) from the impedance - phase relationship \( \cos\phi=\frac{R}{Z} \)).

Answer:

\( i_{rms}^2R \)