Q.An electrical device draws $2\ \text{kW}$ power from AC mains (voltage $223\ \text{V}$ (rms) $= \sqrt{50000}\ \text{V}$). The current differs (lags) in phase by $\phi$ $\left(\tan\phi = -\dfrac{3}{4}\right)$ as compared to voltage. Find

Concept: Power Dissipation in an AC circuit — $P=I_{\text{rms}}^2R=V_{\text{rms}}I_{\text{rms}}\cos\phi$, using the exact given $V_{\text{rms}}=\sqrt{50000}=100\sqrt5\ \text{V}$ (not the rounded $223\ \text{V}$).

1. $\tan\phi=-\dfrac34\Rightarrow\cos\phi=\dfrac45$.
2. $I_{\text{rms}}=\dfrac{P}{V_{\text{rms}}\cos\phi}=\dfrac{2000}{100\sqrt5\times\frac45}=5\sqrt5\approx11.18\ \text{A}$.
3. $R=\dfrac{P}{I_{\text{rms}}^2}=\dfrac{2000}{125}=16\ \Omega$.
4. $X_C-X_L=R\tan\phi=16\times\left(-\dfrac34\right)=-12\ \Omega$.
5. $I_M=\sqrt2\,I_{\text{rms}}=5\sqrt{10}\approx15.81\ \text{A}$. …

Why this approach works

The real power drawn by an AC circuit is dissipated only in its resistive part; the phase lag $\phi$ between current and voltage fixes the ratio of net reactance to resistance. Given the power, the rms voltage, and $\tan\phi$, all three quantities $R$, $X_C-X_L$, and the peak current $I_M$ follow directly.

Since the current lags the voltage, the circuit is net inductive; the given $\tan\phi=-\dfrac{3}{4}$ carries this sign directly ($X_C-X_L$ is negative because $X_L>X_C$).

Step-by-step solution

1. Find $\cos\phi$ from $\tan\phi$.

2. Find $I_{\text{rms}}$ from the power equation.

Using the exact given value $V_{\text{rms}}=\sqrt{50000}\ \text{V}=100\sqrt5\ \text{V}$ (rather than the rounded $223\ \text{V}$):

3. Find $R$.

4. Find $X_C-X_L$.

5. Find the peak current $I_M$.

Effect of doubling R, $X_C$, $X_L$

If a second device has $R'=2R=32\ \Omega$, $X_C'=2X_C$, $X_L'=2X_L$, then $X_C'-X_L'=2(X_C-X_L)=-24\ \Omega$.

The new phase angle is

$\tan\phi'=\frac{X_C'-X_L'}{R'}=\frac{2(X_C-X_L)}{2R}=\tan\phi,$ …

Method: Solving AC "Power Triangle" Problems (Given Power, RMS Voltage, and Phase Angle)

This method applies to any question that gives the power drawn, the rms supply voltage, and the phase angle (or $\tan\phi$) of an AC circuit, and asks for the resistance, net reactance, and/or peak current — plus how scaling every element affects the result.

Steps

Step 1: Get $\cos\phi$ from $\tan\phi$

If given $\tan\phi = a/b$, use the right-triangle relationship (hypotenuse $=\sqrt{a^2+b^2}$):

Keep the sign convention in mind: check whether the question describes current lagging or leading before assigning signs to the reactance terms later.

Step 2: Find $I_{\text{rms}}$ from the real-power equation

Use the EXACT given value of $V_{\text{rms}}$ (not a rounded version) if the problem supplies one, to avoid compounding rounding error through the rest of the solution.

Step 3: Find $R$ from $P = I_{\text{rms}}^2 R$

This works because only the resistive part of the circuit actually dissipates real power.

Step 4: Find the net reactance and the peak current …