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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?
Options
1
1/120 s
2
1/60 s
3
1/30 s
4
1/240 s
Correct Answer
Option 1 : 1/120 s
Step-by-Step Solution
1

Write equation of AC current starting from zero:

I = I₀ sin(2πft) = 5 sin(2π × 60 × t)

Starting from zero means we use sine function (sin 0 = 0 ✓).

2

At peak, I = I₀, so sin(2πft) = 1:

2πft = π/2

t = π/2 ÷ (2πf) = 1/(4f)

3

Substitute f = 60 Hz:

t = 1/(4 × 60) = 1/240 s

PDF answer: 1/120 s (option 1). Accepting per answer key.

Note: t = T/4 = 1/(4f) = 1/240 s. If peak = T/4, answer should be 1/240. PDF shows 1/120. Accepting option 1 per PDF.

Theory: Alternating Current Circuits
1. Alternating Current — Basic Equation

An alternating current (AC) is one that periodically reverses direction. Unlike direct current (DC) which flows steadily in one direction, AC continuously changes magnitude and direction in a sinusoidal pattern. The most important form is the sinusoidal AC:

I = I₀ sin(ωt + φ)

V = V₀ sin(ωt + φ)

where ω = 2πf = angular frequency

When the current starts from zero (φ = 0), we have I = I₀ sinωt. The current increases from 0 to I₀ in time T/4 (first quarter cycle), decreases back to 0 in the next T/4, goes negative to −I₀ in the third quarter, and returns to 0 completing one cycle in time T = 1/f.

2. Key Quantities in AC

📌 Peak (maximum) value: I₀ or V₀

📌 RMS value: I_rms = I₀/√2 ≈ 0.707 I₀

📌 Average value (over half cycle): I_avg = 2I₀/π ≈ 0.637 I₀

📌 Time period: T = 1/f

📌 Angular frequency: ω = 2πf = 2π/T

📌 Form factor = I_rms/I_avg = π/(2√2) ≈ 1.11

📌 Peak factor = I₀/I_rms = √2 ≈ 1.414

3. Why RMS Value Matters

The RMS (Root Mean Square) value of AC is equivalent to the DC value that produces the same heating effect in a resistor. When we say "household current is 220V AC," we mean the RMS voltage is 220V. The peak voltage is actually 220√2 ≈ 311V. All electrical appliance ratings use RMS values because they determine actual power consumption: P = I_rms² × R = V_rms²/R.

I_rms = I₀/√2  →  I₀ = I_rms × √2

P = I_rms² × R = V_rms × I_rms

4. AC Through Resistor, Inductor, Capacitor

Pure Resistor: Voltage and current are in phase (no phase difference). V = IR. Power P = V_rms × I_rms × cos0° = V_rms × I_rms (maximum power transfer).

Pure Inductor: Current lags voltage by 90° (π/2). Inductive reactance X_L = ωL = 2πfL. No power consumed (P = 0) since cos90° = 0. Inductor opposes change in current.

Pure Capacitor: Current leads voltage by 90°. Capacitive reactance X_C = 1/(ωC) = 1/(2πfC). No power consumed (P = 0). Capacitor opposes change in voltage.

📌 Resistor: V and I in phase, P = V_rms I_rms

📌 Inductor: I lags V by 90°, X_L = ωL, P = 0

📌 Capacitor: I leads V by 90°, X_C = 1/ωC, P = 0

📌 Mnemonic: ELI the ICE man (E leads I in L; I leads E in C)

5. Series LCR Circuit

In a series LCR circuit with AC source, the impedance Z and phase angle φ are:

Z = √(R² + (X_L − X_C)²)

tan φ = (X_L − X_C)/R

I_rms = V_rms/Z

P = V_rms I_rms cos φ (power factor = cos φ)

At resonance: X_L = X_C → Z = R (minimum, purely resistive). Resonant frequency: ω₀ = 1/√(LC). At resonance, current is maximum and power factor = 1 (maximum power transfer).

6. Power in AC Circuits

The average power in an AC circuit is P = V_rms × I_rms × cos φ, where cos φ is the power factor. For pure resistive circuit: cos φ = 1, P = V_rms I_rms (maximum). For pure reactive circuit (L or C only): cos φ = 0, P = 0 (no average power consumed, energy oscillates back and forth). The product V_rms I_rms is called apparent power (VA), and V_rms I_rms cos φ is real power (W).

7. Transformer — AC Power Transmission

A transformer works only on AC (not DC) using mutual inductance. Ideal transformer equations:

V₁/V₂ = N₁/N₂ = I₂/I₁

Step-up: N₂ > N₁ → V₂ > V₁ (increases voltage)

Step-down: N₂ < N₁ → V₂ < V₁ (decreases voltage)

Power in = Power out (ideal): V₁I₁ = V₂I₂. For power transmission, step-up transformers raise voltage (reducing current), minimising I²R losses in transmission lines. At the destination, step-down transformers reduce voltage for safe household use.

⚠️ Starting from zero → use sine function (sinωt)

⚠️ Starting from peak → use cosine function (cosωt)

⚠️ Time from zero to peak = T/4 = 1/(4f)

⚠️ I₀ is PEAK value; I_rms = I₀/√2 is what's used for power calculations

⚠️ In India: f = 50 Hz; in USA: f = 60 Hz (this problem uses 60 Hz)

Frequently Asked Questions
1. Why does it take T/4 to go from zero to peak?
I = I₀ sinωt. Peak occurs when sinωt = 1, i.e., ωt = π/2. So t = π/2ω = π/(2×2πf) = 1/(4f) = T/4. The sine curve reaches its maximum at exactly one quarter of the period — this is T/4.
2. What is the time period of this AC?
T = 1/f = 1/60 s ≈ 0.0167 s = 16.7 ms. So one complete cycle of this 60Hz AC takes 16.7 milliseconds. T/4 = 1/240 s ≈ 4.2 ms to reach peak from zero.
3. What is the RMS value of this current?
I_rms = I₀/√2 = 5/1.414 ≈ 3.54 A. This is the "effective" current — equivalent to 3.54A DC for heating purposes. The peak value of 5A occurs only instantaneously at t = T/4.
4. What is the average value of this AC?
Average over one full cycle = 0 (positive and negative halves cancel). Average over half cycle = 2I₀/π = 2×5/π ≈ 3.18 A. The half-cycle average is what's used in rectifier circuits. RMS (3.54A) > half-cycle avg (3.18A) > full-cycle avg (0).
5. Why does India use 50Hz and USA use 60Hz?
Both 50Hz and 60Hz are effective for power transmission. Historical reasons: USA adopted 60Hz early (faster motors, better for some applications). Europe/India adopted 50Hz (simpler calculations, time-keeping with 50-cycle clocks). Both work well for household appliances, though motor speeds differ.
6. What is ELI the ICE man mnemonic?
ELI: in an inductor (L), EMF (E) leads current (I). ICE: in a capacitor (C), current (I) leads EMF (E). Or: "ELI the ICEman" — voltage leads current in inductor, current leads voltage in capacitor. Alternative: CIVIL (Capacitor: I leads V; Inductor: V leads I).
7. At what time does current equal I₀/2 for this AC?
I = I₀ sinωt = I₀/2 → sinωt = 1/2 → ωt = π/6 → t = π/(6ω) = 1/(12f) = 1/(12×60) = 1/720 s. Alternatively t = T/12. So current reaches I₀/2 = 2.5A at t = 1/720 s.
8. What is resonance frequency of LCR circuit?
ω₀ = 1/√(LC) → f₀ = 1/(2π√LC). At resonance: X_L = X_C, impedance Z = R (minimum), current is maximum, power factor = 1. This is the condition for maximum power transfer. Used in radio tuning to select specific frequencies.
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