Q.31 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 31 Solution

Topic: Alternating Current · Official NTA Booklet Code M1

Question

An inductor of inductance L, a capacitor of capacitance C and a resistor of resistance 'R' are connected in series to an ac source of potential difference 'V' volts as shown in figure. Potential difference across L, C and R is 40 V, 10 V and 40 V, respectively. The amplitude of current flowing through LCR series circuit is 10 2 A. The impedance of the circuit is:

✓

Verified Against NTA Final Answer Key

Option (4) : 5 Ω

📚 Alternating Current Easy

📋 Given Parameters & Problem Setup

Inductor Voltage (V_L) 40 V

Capacitor Voltage (V_C) 10 V

Resistor Voltage (V_R) 40 V

Current Amplitude (I_0) 10√2 A

⚡

The 30-Second Exam Hall Shortcut: ⚡ Phasor Triangle: 40 - 10 = 30; √(40² + 30²) = 50 V; Z = 50/10 = 5 Ω

⏱️ Save ~25 seconds

Net reactive voltage $V_L - V_C = 40 - 10 = 30\text{ V}$ . Source RMS voltage $V = \sqrt{40^2 + 30^2} = 50\text{ V}$ (classic 3-4-5 right triangle!). Peak current is $10\sqrt{2}\text{ A}$ , so RMS current is $I_{\text{rms}} = 10\text{ A}$ . Impedance $Z = V/I = 50 / 10 = 5\ \Omega$ ! Match Option (4)!

1 Compute net reactive voltage:

V_X = |V_L - V_C| = 40\text{ V} - 10\text{ V} = 30\text{ V}

2 Total RMS voltage across circuit:

V_{\text{rms}} = \sqrt{V_R^2 + V_X^2} = \sqrt{40^2 + 30^2} = 50\text{ V}

3 Convert current amplitude to RMS current:

I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10\sqrt{2}\text{ A}}{\sqrt{2}} = 10\text{ A}

4 Compute impedance:

Z = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{50\text{ V}}{10\text{ A}} = 5\ \Omega

5 Match Option (4).

🛡️

Why You Can 100% Trust This Shortcut: Canonical AC phasor diagram relations for series RLC circuits and Ohm's law impedance definition.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Calculating the Total Applied RMS Source Voltage

In a series LCR alternating current circuit, the voltages across the inductor and capacitor are $180^\circ$ out of phase with each other, while the voltage across the resistor is in phase with current. Using the phasor voltage addition theorem:

V = \sqrt{V_R^2 + (V_L - V_C)^2} = \sqrt{40^2 + (40 - 10)^2} = \sqrt{40^2 + 30^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50\text{ V}

Step 2

Converting Current Amplitude to RMS Current

The problem statement specifies that the current amplitude (peak value $I_0$ ) is $10\sqrt{2}\text{ A}$ . The corresponding RMS current is:

I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10\sqrt{2}\text{ A}}{\sqrt{2}} = 10\text{ A}

Step 3

Determining the Circuit Impedance (Z)

The impedance $Z$ of the LCR series circuit is defined as the ratio of total RMS voltage to total RMS current (or peak voltage $V_0 = 50\sqrt{2}\text{ V}$ to peak current $I_0$ ):

Z = \frac{V_{\text{rms}}}{I_{\text{rms}}} = \frac{50\text{ V}}{10\text{ A}} = 5\ \Omega

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 1

✗ Incorrect

Verdict & Scientific Reason:

Result of using only the resistor voltage with peak current ( $40 / 10\sqrt{2}$ without inverting $\sqrt{2}$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Ignoring reactive elements in the circuit.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Result of dividing RMS voltage (50 V) directly by peak current ( $10\sqrt{2}\text{ A}$ ) without converting to peak voltage.

⚠️

Student Fallacy (Why candidates pick this trap): Dividing RMS voltage by peak current without matching RMS/peak representations.

3 3

✗ Incorrect

Verdict & Scientific Reason:

This is the pure resistance $R = V_R / I_{\text{rms}} = 40 / 10 = 4\ \Omega$ , not the total circuit impedance $Z$ .

⚠️

Student Fallacy (Why candidates pick this trap): Confusing resistance

R

with total impedance

Z

4 4

✓ 100% Correct

Verdict & Scientific Reason:

Exact total impedance: $Z = V_{\text{rms}} / I_{\text{rms}} = 50\text{ V} / 10\text{ A} = 5\ \Omega$ .

💡

Why this is airtight: None; rigorous AC phasor and RMS calculation.

⚠️

NTA Trap Alert: Current Amplitude vs. RMS Current Trap

The phrase "amplitude of current" denotes peak current $I_0 = 10\sqrt{2}\text{ A}$ , NOT RMS current. Dividing $V = 50\text{ V}$ by $10\sqrt{2}\text{ A}$ produces $5/\sqrt{2}\ \Omega$ , which is Option (2). One must divide either RMS by RMS ( $50/10 = 5$ ) or Peak by Peak ( $50\sqrt{2}/10\sqrt{2} = 5$ ).

🛡️ How to avoid the -1 mark: Check the wording: "amplitude" = peak.

I_{\text{rms}} = I_0 / \sqrt{2} = 10\text{ A}

📖

Core NEET Syllabus Standard

NCERT Physics · Alternating Current Standard Benchmark

Core Scientific Principle Tested:

"Impedance is the ratio of effective voltage to effective current: $Z = \frac{V}{I} = \sqrt{R^2 + (X_L - X_C)^2}$ ."

Syllabus Focus: Series LCR Circuit and Impedance

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 31 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Phasor Triangle: 40 - 10 = 30; √(40² + 30²) = 50 V; Z = 50/10 = 5 Ω

Step-by-Step Rigorous Workflow & Mathematical Proof

Calculating the Total Applied RMS Source Voltage

Converting Current Amplitude to RMS Current

Determining the Circuit Impedance (Z)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Current Amplitude vs. RMS Current Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Alternating Current

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