A resistance `R` draws power `P` when connected to an `AC` source. If an inductance is now placed in series with the resistance, such that the impedence of the circuit becomes `Z`, the power drawn will be

Options

A. `P(R/Z)^(2)`
B. `Psqrt(R/Z)`
C. `P(R/Z)`
D. `P`

Explanation

To solve the problem, we need to determine the power drawn by a circuit when a resistance \( R \) is connected to an AC source, and then an inductance is added in series with the resistance, resulting in an impedance \( Z \).

### Step-by-Step Solution:

1. **Power in Pure Resistance**:
When the resistance \( R \) is connected to an AC source, the power \( P \) drawn by the circuit can be expressed using the formula:
\[ P = \frac{V^2}{R} \]
where \( V \) is the RMS voltage across the resistance.

2. **Calculate RMS Voltage**:
From the power formula, we can rearrange to find the RMS voltage:
\[ V^2 = P \cdot R \]
This equation will help us later when we consider the circuit with the inductor.

3. **Adding Inductance in Series**:
When an inductance \( L \) is added in series with the resistance \( R \), the total impedance \( Z \) of the circuit is given by:
\[ Z = \sqrt{R^2 + X_L^2} \]
where \( X_L = \omega L \) is the inductive reactance.

4. **Power in RL Circuit**:
The power \( P' \) drawn by the RL circuit can be expressed as:
\[ P' = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi \]
where \( \cos \phi = \frac{R}{Z} \) is the power factor.

5. **Expressing Current in Terms of Voltage and Impedance**:
The current \( I_{\text{rms}} \) can be expressed as:
\[ I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} \]

6. **Substituting Values**:
Substituting \( I_{\text{rms}} \) and \( \cos \phi \) into the power formula gives:
\[ P' = V_{\text{rms}} \cdot \left(\frac{V_{\text{rms}}}{Z}\right) \cdot \left(\frac{R}{Z}\right) \]
Simplifying this, we get:
\[ P' = \frac{V_{\text{rms}}^2 \cdot R}{Z^2} \]

7. **Substituting for \( V_{\text{rms}}^2 \)**:
Now, substituting \( V_{\text{rms}}^2 = P \cdot R \) into the equation for \( P' \):
\[ P' = \frac{(P \cdot R) \cdot R}{Z^2} = \frac{P \cdot R^2}{Z^2} \]

8. **Final Result**:
Thus, the power drawn by the circuit with the inductance in series is:
\[ P' = \frac{P \cdot R^2}{Z^2} \]