AC Power Calculation with Impedance
The initial power P drawn by a resistor R connected to a constant AC voltage source ($V_{rms}$) is:
$P = \frac{V_{rms}^2}{R}$
When an inductance is added in series with the resistor, the total impedance of the circuit becomes Z. The source voltage $V_{rms}$ remains unchanged.
The new RMS current $I'_{rms}$ in this series R-L circuit is calculated using Ohm's law for AC circuits:
$I'_{rms} = \frac{V_{ سےms}}{Z}$
The power dissipated by the resistor in the new circuit is the new power $P'$. Since only the resistor dissipates average power in an AC circuit:
$P' = (I'_{rms})^2 R$
Substitute the expression for $I'_{rms}$:
$P' = \left(\frac{V_{rms}}{Z}\right)^2 R = \frac{V_{rms}^2}{Z^2} R$
From the initial power equation ($P = \frac{V_{rms}^2}{R}$), we can rearrange to find $V_{rms}^2$:
$V_{rms}^2 = P \times R$
Now, substitute this expression for $V_{rms}^2$ into the equation for $P'$:
$P' = \frac{(P \times R)}{Z^2} R$
Simplify the expression:
$P' = P \frac{R^2}{Z^2}$
$P' = P \left(\frac{R}{Z}\right)^2$
Thus, the power drawn by the circuit in the presence of the inductor will be $P \left( \frac{R}{Z} \right)^2$.
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