An electrical device draws 2 kW power form AC mains [voltage 223 V (rms) `= sqrt(50,000) V`]. The current differs (lags) in phase by `phi (tan = (-3)/(4))` as compared to voltage. Find (i) R, (ii) `X_(C ) - X_(L)`, and (iii) `I_(M)`. Another device has twice the values for R, `X_(C )` and `X_(L)`. How are the answers affected ?

Explanation

To solve the problem step by step, we will follow the given information and apply the relevant formulas.

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

### Step 1: Calculate Impedance (Z)
Using the formula for power in an AC circuit:
\[ P = \frac{V^2}{Z} \]
We can rearrange this to find Z:
\[ Z = \frac{V^2}{P} \]
Substituting the values:
\[ Z = \frac{(223)^2}{2000} = \frac{49729}{2000} = 24.8645 \approx 25 \, \Omega \]

### Step 2: Calculate Resistance (R)
We know from the relationship involving tan φ:
\[ \tan φ = \frac{X_C - X_L}{R} \]
Given that \( \tan φ = -\frac{3}{4} \):
\[ -\frac{3}{4} = \frac{X_C - X_L}{R} \]
Thus,
\[ X_C - X_L = -\frac{3}{4} R \]

### Step 3: Use the Impedance Formula
The impedance Z can also be expressed as:
\[ Z^2 = R^2 + (X_L - X_C)^2 \]
Substituting the known value of Z:
\[ 25^2 = R^2 + (X_L - X_C)^2 \]
This gives us:
\[ 625 = R^2 + (X_L - X_C)^2 \]

### Step 4: Substitute for \(X_C - X_L\)
From the equation \(X_C - X_L = -\frac{3}{4} R\), we can express \(X_L - X_C\) as:
\[ X_L - X_C = \frac{3}{4} R \]
Substituting this into the impedance equation:
\[ 625 = R^2 + \left(\frac{3}{4} R\right)^2 \]
Calculating \(\left(\frac{3}{4} R\right)^2\):
\[ \left(\frac{3}{4} R\right)^2 = \frac{9}{16} R^2 \]
So, we have:
\[ 625 = R^2 + \frac{9}{16} R^2 \]
Combining terms:
\[ 625 = \left(1 + \frac{9}{16}\right) R^2 = \frac{25}{16} R^2 \]
Thus:
\[ R^2 = \frac{625 \times 16}{25} = 400 \]
Taking the square root:
\[ R = 20 \, \Omega \]

### Step 5: Calculate \(X_C - X_L\)
Now substituting R back into the equation for \(X_C - X_L\):
\[ X_C - X_L = -\frac{3}{4} \times 20 = -15 \, \Omega \]

### Step 6: Calculate Maximum Current \(I_M\)
The RMS current \(I\) can be calculated as:
\[ I = \frac{V}{Z} = \frac{223}{25} = 8.92 \, A \]
To find the maximum current \(I_M\):
\[ I_M = \sqrt{2} \times I = \sqrt{2} \times 8.92 \approx 12.6 \, A \]

### Summary of Results:
1. Resistance \(R = 20 \, \Omega\)
2. \(X_C - X_L = -15 \, \Omega\)
3. Maximum Current \(I_M \approx 12.6 \, A\)

### Part (ii): Effect of Doubling R, \(X_C\), and \(X_L\)
If another device has twice the values for \(R\), \(X_C\), and \(X_L\):
- New \(R = 40 \, \Omega\)
- New \(X_C - X_L = -30 \, \Omega\)

The impedance \(Z\) will also increase, leading to:
- New \(Z\) will be higher, thus the current will decrease (half of the original).
- Power will also be half since \(P = I^2 R\) and \(I\) is halved.