An electrical device draws 2 kW power from ac mains voltage 223 V(rms). The current differs lags in phase by `phi=tan^(-1)(-(3)/(4))` as compared to valtage. The resistance R in the circuit is

Options

A. `15 Omega`
B. `20 Omega`
C. `25 Omega`
D. `30 Omega`

Explanation

To find the resistance \( R \) in the circuit, we can follow these steps:

### Step 1: Identify Given Values
- Power \( P = 2 \, \text{kW} = 2000 \, \text{W} \)
- RMS Voltage \( V_{\text{rms}} = 223 \, \text{V} \)
- Phase angle \( \phi = \tan^{-1}\left(-\frac{3}{4}\right) \)

### Step 2: Calculate the Impedance \( Z \)
Using the formula for power in an AC circuit:
\[ P = \frac{V_{\text{rms}}^2}{Z} \]
We can rearrange this to find \( Z \):
\[ Z = \frac{V_{\text{rms}}^2}{P} \]
Substituting the known values:
\[ Z = \frac{(223)^2}{2000} \]
Calculating \( Z \):
\[ Z = \frac{49729}{2000} = 24.86 \, \Omega \approx 25 \, \Omega \]

### Step 3: Relate Phase Angle to Impedance
The phase angle \( \phi \) is given by:
\[ \tan \phi = \frac{X_L - X_C}{R} \]
From the given \( \phi \):
\[ \tan \phi = -\frac{3}{4} \]
Thus,
\[ X_L - X_C = -\frac{3}{4} R \]
Let’s call this Equation (1).

### Step 4: Use Impedance Formula
The impedance \( Z \) can also be expressed as:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Substituting \( Z = 25 \, \Omega \) and \( X_L - X_C = -\frac{3}{4} R \):
\[ 25 = \sqrt{R^2 + \left(-\frac{3}{4} R\right)^2} \]
Squaring both sides:
\[ 625 = R^2 + \left(\frac{9}{16} R^2\right) \]
Combining terms:
\[ 625 = R^2 + \frac{9}{16} R^2 = \frac{16R^2 + 9R^2}{16} = \frac{25R^2}{16} \]
Multiplying through by 16:
\[ 10000 = 25R^2 \]
Dividing by 25:
\[ R^2 = 400 \]
Taking the square root:
\[ R = 20 \, \Omega \]

### Conclusion
The resistance \( R \) in the circuit is:
\[ \boxed{20 \, \Omega} \]