Q.154 Physics ⭐ 360° Option Elimination Active

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NEET 2020 Physics Question 154 Solution

Topic: Alternating Current · Official NTA Booklet Code E1

Question

A 40 μF capacitor is connected to a 200 V, 50 Hz ac supply. The rms value of the current in the circuit is, nearly:

✓

Verified Against NTA Final Answer Key

Option (3) : 2.5 A

📚 Alternating Current Easy

📋 Given Parameters & Problem Setup

Capacitance (C) 40 µF = 40 × 10^-6 F

RMS Voltage (V_rms) 200 V

AC Frequency (f) 50 Hz

Formula I_rms = V_rms / X_C = V_rms × 2π f C

⚡

The 30-Second Exam Hall Shortcut: ⚡ "I_rms = V_rms × 2πfC = 200 × (2 × π × 50) × (40 × 10^-6) = 200 × 100π × 40 × 10^-6 = 0.8π ≈ 2.51 A ≈ 2.5 A!"

⏱️ 20 seconds

The capacitive reactance is $X_C = \frac{1}{2\pi f C}$ . The rms current is $I_{\text{rms}} = V_{\text{rms}} \cdot 2\pi f C$ . Substitute values: $2\pi f = 2\pi \times 50 = 100\pi\text{ rad/s}$ . Then $I_{\text{rms}} = 200 \times 100\pi \times 40 \times 10^{-6} = 800000\pi \times 10^{-6} = 0.8\pi\text{ A}$ . Since $\pi \approx 3.1416$ , $0.8 \times 3.1416 = 2.513\text{ A} \approx 2.5\text{ A}$ ! Mark Option (3) in 10 seconds.

1 Formula: I_rms = V_rms × ω C = V_rms × 2π f C.

2 ω = 2π × 50 = 100π rad/s.

3 I_rms = 200 V × 100π s^-1 × 40 × 10^-6 F.

4 I_rms = 200 × 40 × 100π × 10^-6 = 800,000π × 10^-6 = 0.8π A.

5 0.8 × 3.1416 ≈ 2.51 A ≈ 2.5 A.

6 Select Option (3) 2.5 A.

🛡️

Why You Can 100% Trust This Shortcut: NCERT Physics · Alternating Current (AC Voltage Applied to a Capacitor)

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formula for Capacitive Reactance

When an alternating voltage $V = V_0 \sin(\omega t)$ is applied across a pure capacitor of capacitance $C$ , the opposition to current flow is called capacitive reactance ( $X_C$ ): $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$

Step 2

Relate RMS Current to RMS Voltage

By Ohm's law in AC circuits: $I_{\text{rms}} = \frac{V_{\text{rms}}}{X_C} = V_{\text{rms}} \cdot (2\pi f C)$

Step 3

Substitute Given Numerical Values

Given: - $V_{\text{rms}} = 200\text{ V}$ - $f = 50\text{ Hz} \implies \omega = 2\pi \times 50 = 100\pi\text{ rad s}^{-1}$ - $C = 40\text{ }\mu\text{F} = 40 \times 10^{-6}\text{ F}$ $I_{\text{rms}} = 200 \times (100\pi) \times (40 \times 10^{-6})$ $I_{\text{rms}} = 200 \times 4000\pi \times 10^{-6}$ $I_{\text{rms}} = 800000\pi \times 10^{-6} = 0.8\pi\text{ A}$

Step 4

Evaluate Numerical Result

$I_{\text{rms}} = 0.8 \times 3.1416 = 2.51328\text{ A} \approx 2.5\text{ A}$ This matches Option (3).

⭐

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 A

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; arithmetic error.

⚠️

Student Fallacy (Why candidates pick this trap): Dividing by

\sqrt{2}

unnecessarily.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; underestimation of current.

⚠️

Student Fallacy (Why candidates pick this trap): Using 0.65π instead of 0.8π.

3 C

✓ 100% Correct

Verdict & Scientific Reason:

Factually and mathematically exact; $I_{\text{rms}} = 0.8\pi \approx 2.51\text{ A} \approx 2.5\text{ A}$ .

💡

Why this is airtight: None; clean AC capacitive circuit calculation.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; off by a factor of 10 in decimal placement.

⚠️

Student Fallacy (Why candidates pick this trap): Power of 10 error in microfarad conversion (

10^{-5}

instead of

10^{-6}

⚠️

NTA Trap Alert: 2πf = 100π for 50 Hz Supply

In India, AC frequency is 50 Hz, so $\omega = 2\pi \times 50 = 100\pi \approx 314\text{ rad/s}$ . Always multiply $\omega C$ directly with $V$ to get $I$ without inverting twice!

🛡️ How to avoid the -1 mark:

I = V \times 2\pi f C

is much faster than computing

X_C

separately!

📖

Core NEET Syllabus Standard

Physics · Alternating Current Standard Benchmark

Core Scientific Principle Tested:

Syllabus Focus: Capacitive Reactance and RMS Current: I_rms = V_rms / X_C = V_rms × 2πfC

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2020 Physics Question 154 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ "I_rms = V_rms × 2πfC = 200 × (2 × π × 50) × (40 × 10^-6) = 200 × 100π × 40 × 10^-6 = 0.8π ≈ 2.51 A ≈ 2.5 A!"

Step-by-Step Rigorous Workflow & Mathematical Proof

Formula for Capacitive Reactance

Relate RMS Current to RMS Voltage

Substitute Given Numerical Values

Evaluate Numerical Result

360° Option Elimination Engine Key Feature

NTA Trap Alert: 2πf = 100π for 50 Hz Supply

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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