A.C voltage across a 0.5μF capacitor is 16sin(2×10³t)V, peak value of voltage through the capacitor is:
- 2×10³
- 10³
- 16
- 2
Show answer and explanation
Answer: C. 16
Comparing with V=V₀sin(ωt), the peak voltage V₀ is simply the amplitude given: 16 V.
Practise 28 real Federal Board questions on Alternating Current from 2023 to 2026. See the answer and explanation as soon as you answer.
Alternating Current is one of the 14 main chapters in our Federal Board FSC Physics question bank (12th class), and about 6% of its real questions come from it. Tayyari Karo has 28 real questions on this chapter from 2023 to 2026; 8 are multiple choice and marked instantly, and the rest are short and long questions with a model answer. By difficulty: 18 medium, 9 easy, 1 hard.
Work through the questions below, then open the explanation for each one. Redo the ones you missed a few days later.
These are real past-paper questions. Think of your answer first, then open the explanation.
A.C voltage across a 0.5μF capacitor is 16sin(2×10³t)V, peak value of voltage through the capacitor is:
Answer: C. 16
Comparing with V=V₀sin(ωt), the peak voltage V₀ is simply the amplitude given: 16 V.
Primary function of transformer is to change:
Answer: B. The level of alternating voltage
A transformer's main purpose is to step up or step down the magnitude (level) of an alternating voltage via mutual induction.
SI unit of capacitive reactance is:
Answer: A. Ohm
Reactance (like resistance) is measured in ohms (Ω).
If N_s/N_p = 1, then the transformer is:
Answer: A. Neither step up nor step down
Equal turns ratio (N_s = N_p) means V_s = V_p — the output voltage equals the input voltage, i.e. neither stepped up nor down (a 1:1 isolation transformer).
How does doubling the frequency affect the capacitive reactance? Show mathematically.
Capacitive reactance: X_c = 1/(2πfC)
If frequency f is doubled (f→2f):
X_c' = 1/(2π(2f)C) = (1/2)[1/(2πfC)]
X_c' = X_c/2 — doubling the frequency halves the capacitive reactance.
A pure inductor is connected across a 10V, 200Hz AC Supply and the current flowing through it is measured as 0.4A. Determine the value of its inductance.
X_L = V/I = 10/0.4 = 25 Ω
X_L = 2πfL ⇒ L = X_L/(2πf) = 25/(2π×200) = 25/1256.6 ≈ 0.0199 H (≈ 19.9 mH)
In a transformer, why is a laminated iron core used instead of a solid one?
A solid iron core would allow large eddy currents to circulate within its bulk (driven by the changing flux), causing significant I²R heating and wasted energy. Using a laminated core — thin iron sheets electrically insulated from each other (e.g. by a varnish coating) — breaks up the conducting paths available to eddy currents, greatly increasing their resistance and reducing their magnitude, which substantially reduces eddy current (energy) losses and keeps the transformer running efficiently and cooler.
A simple AC generator consists of N=10 turns coil of area A=1200 cm² which rotates at a constant frequency f=60 Hz in a uniform magnetic field B=0.40 T. Find the peak e.m.f generated by it.
ε_max = NBAω = NBA(2πf)
A = 1200 cm² = 1200×10⁻⁴ m² = 0.12 m²
ε_max = 10 × 0.40 × 0.12 × 2π × 60 = 0.48 × 376.99 ≈ 181 V
| Session | Questions |
|---|---|
| 2026 Annual | 5 |
| 2025 1st Annual | 7 |
| 2025 2nd Annual | 4 |
| 2024 Annual | 7 |
| 2023 Annual | 5 |
28 real Federal Board (FBISE) Physics questions on Alternating Current from 2023 to 2026, for 12th class. Each one shows the session it came from.
No. We could not find an official FBISE marking scheme for these papers, so the answers and explanations are written by Tayyari Karo for practice. Check anything important with your teacher or textbook.
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