Step 1: Understanding the Concept:
In an AC circuit, average power is only dissipated by the resistive part of the circuit. Inductors and capacitors do not consume average power over a cycle; they only store and release energy.
The power dissipated in a circuit with a resistor and an inductor in series is given by \(P = I_{rms}^2 R\).
When the inductor is added, the total impedance of the circuit increases, which reduces the current drawn from the source.
The power factor of the circuit also changes. The power factor is defined as \(\cos \phi = R/Z\).
Step 2: Key Formula or Approach:
Initial state (Pure resistor): \(P = \frac{V_{rms}^2}{R}\).
Final state (Series R-L): \(P_{new} = I_{rms}^2 R\), where \(I_{rms} = \frac{V_{rms}}{Z}\).
Step 3: Detailed Explanation:
Let the voltage of the AC source be \(V_{rms}\).
Initially, only the resistor \(R\) is connected. The power drawn is:
\[ P = \frac{V_{rms}^2}{R} \]
From this, we can express the square of the source voltage as:
\[ V_{rms}^2 = P \cdot R \]
Now, an inductor is added in series. The impedance of this new R-L circuit is given as \(Z\).
The RMS current in the new circuit is:
\[ I'_{rms} = \frac{V_{rms}}{Z} \]
The power drawn in the new circuit, \(P_{new}\), is still only dissipated across the resistor:
\[ P_{new} = (I'_{rms})^2 \cdot R \]
Substitute the expression for \(I'_{rms}\):
\[ P_{new} = \left( \frac{V_{rms}}{Z} \right)^2 \cdot R = \frac{V_{rms}^2 R}{Z^2} \]
Now, substitute the value of \(V_{rms}^2\) we found from the initial state (\(V_{rms}^2 = P \cdot R\)):
\[ P_{new} = \frac{(P \cdot R) \cdot R}{Z^2} = \frac{P \cdot R^2}{Z^2} \]
\[ P_{new} = P \left( \frac{R}{Z} \right)^2 \]
This matches option (C).
Step 4: Final Answer:
The power drawn is proportional to the square of the RMS current. Since the current decreases by a factor of \(R/Z\), the power decreases by the square of that factor, resulting in \(P(R/Z)^2\).