NEET · Physics —

Alternating Current appeared 2 times across 1 year — 4.4% of Physics. This question is from AC Voltage Applied to a Resistor.

Year 2024 Total
Questions 2 2
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The peak value of an alternating current is 5 ~A and frequency is 60 ~Hz. How long will the current, starting from zero, take to reach the peak value?

Solution & Explanation

Related Formula
i = ipeak (ω t)

ω = 2π f

Core Logic

Given: Peak current ipeak = 5 ~A Frequency f = 60 ~Hz Angular frequency ω = 2π × 60 = 120π ~rad/s

The equation of the alternating current is:

i = 5 (120π t)

We need to find the time t when the current i reaches its peak value (5 ~A) for the first time.

Step 1: Calculate Time
5 = 5 (120π t) (120π t) = 1

The smallest positive angle for which sine is 1 is (π)/(2):

120π t = (π)/(2) t = (1)/(120 × 2) t = (1)/(240) ~s
Pattern Recognition

An AC wave starting from zero reaches its first peak exactly at a quarter of a full time period (T/4). Since T = 1/f = 1/60 ~s, the quarter period is 1 / (4 × 60) = 1/240 ~s.

Chapter Mix

Class 12 Physics: Alternating Current

Reference Study Guides

Alternating Current

More Alternating Current Previous-Year Questions

Q7 neet_2026_03_may_morning Resonance in Series LCR Circuit
An ac circuit contains a resistance of 1 ~kΩ, a capacitor of 0.1 ~μ F and an inductor of 1 ~mH connected in series. The resonance frequency of the circuit is approximately.
  • A. 10.1 ~kHz
  • B. 20.7 ~kHz
  • C. 15.9 ~kHz
  • D. 13.5 ~kHz

Solution

Related Formula
f₀ = 12π √(LC)
Core Logic

At resonance, the frequency depends only on the inductance (L) and capacitance (C), ignoring resistance (R).

Given: L = 1 ~mH = 10⁻³ ~H C = 0.1 ~μ F = 10⁻⁷ ~F

Step 1: Calculate Frequency
f₀ = 12π 10⁻³ × 10⁻⁷ f₀ = 12π 10⁻¹⁰ f₀ = 12π × 10⁻⁵ f₀ = (10⁵)/(2π) = (100,000)/(6.28) ≈ 15923 ~Hz f₀ ≈ 15.9 ~kHz
Pattern Recognition

Resistance in a series LCR circuit affects the quality factor (sharpness of resonance) but does not shift the natural resonant frequency f₀.

Chapter Mix

Class 12 Physics: Alternating Current

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