Q.A sinusoidal voltage of peak value $283\ \text{V}$ and frequency $50\ \text{Hz}$ is applied to a series LCR circuit in which $R = 3\ \Omega$, $L = 25.48\ \text{mH}$, and $C = 796\ \mu\text{F}$. Find

Concept: Power Dissipation in Resistors — in an AC circuit, only the resistor dissipates power; the average power is $P = V_{\text{rms}} I_{\text{rms}} \cos\phi = I_{\text{rms}}^2 R$.

Step 1 — Reactances and impedance

Inductive reactance:

$X_L = 2\pi f L = 2\pi (50)(25.48 \times 10^{-3}) = 8\ \Omega$

Capacitive reactance:

$X_C = \frac{1}{2\pi f C} = \frac{1}{2\pi (50)(796 \times 10^{-6})} = 4\ \Omega$

Net reactance: $X = X_L - X_C = 4\ \Omega$

Impedance: $Z = \sqrt{R^2 + X^2} = \sqrt{3^2 + 4^2} = 5\ \Omega$

Step 2 — Phase difference and power factor

$\tan\phi = \frac{X}{R} = \frac{4}{3} \Rightarrow \phi = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ$ (voltage leads current)

Power factor: $\cos\phi = \frac{R}{Z} = \frac{3}{5} = 0.6$

Step 3 — Power dissipated

RMS voltage: $V_{\text{rms}} = \frac{283}{\sqrt{2}} \approx 200\ \text{V}$

RMS current: $I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{200}{5} = 40\ \text{A}$

Power: $P = I_{\text{rms}}^2 R = (40)^2 (3) = 4800\ \text{W}$

Concept and Intuition

In a series LCR circuit, the resistor, inductor, and capacitor each oppose current in different ways. Resistance $R$ dissipates energy as heat. Inductive reactance $X_L = \omega L$ and capacitive reactance $X_C = 1/(\omega C)$ store and release energy but do not dissipate it — they merely cause a phase shift between voltage and current.

The total opposition to current is impedance $Z$, given by:

The phase difference $\phi$ tells us whether the circuit behaves more like an inductor (voltage leads current, $\phi > 0$) or a capacitor (current leads voltage, $\phi < 0$):

Power is only dissipated in the resistor. The average power over a cycle is:

where $\cos\phi$ is the power factor.

Step-by-Step Solution

1. Find the angular frequency $\omega$

Given frequency $f = 50\ \text{Hz}$:

2. Calculate inductive reactance $X_L$

$L = 25.48\ \text{mH} = 25.48 \times 10^{-3}\ \text{H}$

Using $\pi \approx 3.14$:

3. Calculate capacitive reactance $X_C$

$C = 796\ \mu\text{F} = 796 \times 10^{-6}\ \text{F}$

First compute $\omega C = 100\pi \times 796 \times 10^{-6} = 314 \times 796 \times 10^{-6}$

$314 \times 796 \approx 250,000$ (since $314 \times 800 = 251,200$, minus $314 \times 4 = 1,256$ gives $249,944$)

So $\omega C \approx 0.25$

4. Compute net reactance $X$

The circuit is inductive (positive reactance).

5. Find impedance $Z$

6. Determine phase difference $\phi$

Since $X_L > X_C$, voltage leads current by $53.13^\circ$.

7. Calculate rms values of source voltage and current

Peak voltage $V_0 = 283\ \text{V}$

8. Find power factor

9. Compute power dissipated

Alternatively, since only the resistor dissipates power:

Method: Phasor Analysis of Series LCR Circuit

This method uses phasor diagrams and impedance triangle to solve AC circuit problems step-by-step.

Step 1: Find Inductive and Capacitive Reactance

Given:

• $V_0 = 283\ \text{V}$, $f = 50\ \text{Hz}$
• $R = 3\ \Omega$, $L = 25.48\ \text{mH} = 25.48 \times 10^{-3}\ \text{H}$
• $C = 796\ \mu\text{F} = 796 \times 10^{-6}\ \text{F}$

Angular frequency:

Inductive reactance:

Capacitive reactance:

Step 2: Calculate Impedance (Part a)

Net reactance:

Impedance magnitude:

Step 3: Find Phase Difference (Part b)

Phase angle $\phi$ (voltage leads current if $X_L > X_C$):

Since $X_L > X_C$, voltage leads current by $53.13^\circ$.

Step 4: Compute Power Dissipated (Part c)

RMS voltage:

RMS current:

Power dissipated (only in resistor):

Step 5: Determine Power Factor (Part d)

Power factor:

The power factor is lagging because the circuit is inductive ($X_L > X_C$).

Quick Verification

• $P = V_{\text{rms}} I_{\text{rms}} \cos \phi = 200 \times 40 \times 0.6 = 4800\ \text{W}$ ✓

Final Answers:

• (a) $Z = 5\ \Omega$
• (b) $\phi = 53.13^\circ$ (voltage leads current)
• (c) $P = 4800\ \text{W}$
• (d) $\cos \phi = 0.6$ (lagging)

Here are the common mistakes students make when solving this exact problem, along with how to avoid each.

Mistake 1: Forgetting to convert units (mH, μF → H, F)

The mistake:

Plugging $L = 25.48$ and $C = 796$ directly into formulas without converting to henries and farads.

How to avoid:

Always write the conversion step explicitly:

• $L = 25.48\ \text{mH} = 25.48 \times 10^{-3}\ \text{H}$
• $C = 796\ \mu\text{F} = 796 \times 10^{-6}\ \text{F}$

Check: If you get an impedance near $3\ \Omega$, you likely converted correctly. If it’s huge or tiny, re-check units.

Mistake 2: Using peak voltage ($V_0$) in RMS formulas for power

The mistake:

Using $P = \frac{V_0^2}{R}$ or $P = V_0 I_0 \cos\phi$ directly — these give peak power, not average power.

How to avoid:

Remember: Power dissipation in AC circuits uses RMS values.

• $V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{283}{\sqrt{2}} \approx 200\ \text{V}$
• Average power: $P = V_{\text{rms}} I_{\text{rms}} \cos\phi$ or $P = I_{\text{rms}}^2 R$

Key fact: Only $I_{\text{rms}}^2 R$ gives the correct average power dissipated.

Mistake 3: Confusing phase difference sign ($\phi$)

The mistake:

Writing $\phi = \tan^{-1}\left(\frac{X_L - X_C}{R}\right)$ but then using the wrong sign when calculating power factor.

How to avoid:

• $X_L = \omega L$, $X_C = \frac{1}{\omega C}$
• If $X_L > X_C$, $\phi > 0$ (voltage leads current — inductive circuit)
• If $X_L < X_C$, $\phi < 0$ (voltage lags current — capacitive circuit)
• Power factor $\cos\phi$ is always positive (use $|\phi|$ or $\cos\phi = \frac{R}{Z}$ directly)

Tip: Use $\cos\phi = \frac{R}{Z}$ — it’s foolproof and avoids sign errors.

Mistake 4: Forgetting $\omega = 2\pi f$ (not $f$)

The mistake:

Using $f = 50\ \text{Hz}$ directly in $X_L = \omega L$ as $X_L = fL$.

How to avoid:

Always write:

Then:

Mistake 5: Calculating impedance $Z$ incorrectly

The mistake:

Writing $Z = R + (X_L - X_C)$ or $Z = \sqrt{R^2 + X_L^2 + X_C^2}$.

How to avoid:

The correct formula is:

Why: $X_L$ and $X_C$ are opposite in phase — they subtract, not add.

Mistake 6: Using $P = V_{\text{rms}} I_{\text{rms}}$ without $\cos\phi$

The mistake:

Assuming $P = V_{\text{rms}} I_{\text{rms}}$ gives power dissipated.

How to avoid:

In an LCR circuit, voltage and current are out of phase. The true power is:

Only the resistive component dissipates power.

Alternative (safer):

This automatically accounts for phase — no $\cos\phi$ needed.

Mistake 7: Rounding too early

The mistake:

Rounding intermediate values (e.g., $X_L$, $X_C$, $Z$) to 2–3 digits, then getting a final answer that’s off.

How to avoid:

Keep at least 4 significant figures in intermediate steps. Round only the final answer.

Example:

• $X_L = 100\pi \times 0.02548 \approx 8.004\ \Omega$ (not $8.0$)
• $X_C = \frac{1}{100\pi \times 796 \times 10^{-6}} \approx 4.000\ \Omega$ (not $4.0$)
• Then $X_L - X_C = 4.004\ \Omega$, $Z = \sqrt{3^2 + 4.004^2} \approx 5.00\ \Omega$

Quick Summary Checklist

Final tip: For part (c), the cleanest path is:

1. Find $Z$
2. $I_{\text{rms}} = V_{\text{rms}} / Z$
3. $P = I_{\text{rms}}^2 R$

This avoids any phase sign confusion and gives the correct answer every time.