An electrical device draws `2 KW` power from `(AC)` mains (voltage `223 V ms=sqrt(50,000) V) .` The current differs (lags) in phase by `phi(tan phi=(-3)/(4))` as compared to voltafge. Find `X_(L)-X_(c)` (in ohm).

Answer

To solve the problem, we need to find the difference between the inductive reactance (X_L) and the capacitive reactance (X_C) given the power, voltage, and phase difference in an AC circuit.

### Step-by-Step Solution:

1. **Given Data:**
- Power (P) = 2 kW = 2000 W
- RMS Voltage (V_RMS) = 223 V
- Phase difference (tan φ) = -3/4

2. **Calculate Impedance (Z):**
The formula for power in an AC circuit is given by:
\[ P = \frac{V^2}{Z} \]
Rearranging this gives us:
\[ Z = \frac{V^2}{P} \]
Substituting the values:
\[ Z = \frac{(223)^2}{2000} \]
\[ Z = \frac{49729}{2000} = 24.8645 \, \text{ohm} \approx 25 \, \text{ohm} \]

3. **Calculate Resistance (R):**
The phase angle (φ) can be related to the impedance components as follows:
\[ \tan φ = \frac{X_C - X_L}{R} \]
From the given tan φ:
\[ \tan φ = -\frac{3}{4} \]
Therefore:
\[ \frac{X_C - X_L}{R} = -\frac{3}{4} \]
Rearranging gives:
\[ X_C - X_L = -\frac{3}{4} R \]

4. **Calculate R using the impedance:**
We know that:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
We can express R in terms of Z and the other reactances. We will assume R is a part of the impedance:
\[ R = Z \cdot \cos φ \]
To find cos φ, we can use:
\[ \cos φ = \frac{1}{\sqrt{1 + \tan^2 φ}} = \frac{1}{\sqrt{1 + \left(-\frac{3}{4}\right)^2}} = \frac{1}{\sqrt{1 + \frac{9}{16}}} = \frac{1}{\sqrt{\frac{25}{16}}} = \frac{4}{5} \]
Now substituting back:
\[ R = Z \cdot \cos φ = 25 \cdot \frac{4}{5} = 20 \, \text{ohm} \]

5. **Calculate X_L - X_C:**
Now we can substitute R back into our equation:
\[ X_C - X_L = -\frac{3}{4} \cdot 20 = -15 \, \text{ohm} \]
Therefore:
\[ X_L - X_C = 15 \, \text{ohm} \]

### Final Answer:
\[ X_L - X_C = 15 \, \text{ohm} \]