Q.14 An alternating voltage source given by e=140sin(314.2 t) is connected across a pure resistor of 50 Ω . Determine : (i) Frequency of the source. (ii) RMS value of emf.

Answer: (i) 50 Hz (ii) ≈ 99 V Comparing with e=e₀sinω t : e₀=140 V, ω=314.2 rad/s. (i) f=ω/2π=314.2/2×3.142=50 Hz. (ii) e_rms=e₀/sqrt2=140/1.414=70sqrt2≈98.99≈99 V. (The rms current, if needed, is I_rms=e_rms/R=98.99/50≈1.98 A.)

An alternating voltage source given by \(e=140\sin(314.2\,t)\) is connected across a pure resistor of 50 \(\Omega\). Determine :(i) Frequency of the source.(ii) RMS value of emf.

Q.14 An alternating voltage source given by \(e=140\sin(314.2\,t)\) is connected across a pure resistor of 50 \(\Omega\). Determine :
(i) Frequency of the source.
(ii) RMS value of emf.

Doubtly Asked 28 Oct 2026 Board exam · June 2026 hsc 12th Save WhatsApp Report

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Answer: (i) 50 Hz (ii) \(\approx 99\) V

Comparing with \(e=e_0\sin\omega t\): \(e_0=140\) V, \(\omega=314.2\) rad/s.

(i) \(f=\dfrac{\omega}{2\pi}=\dfrac{314.2}{2\times3.142}=50\) Hz.

(ii) \(e_{rms}=\dfrac{e_0}{\sqrt2}=\dfrac{140}{1.414}=70\sqrt2\approx98.99\approx99\) V.

(The rms current, if needed, is \(I_{rms}=e_{rms}/R=98.99/50\approx1.98\) A.)

Doubtly AI · 28 Oct 2026 AI-verified · Oct 2026 Report

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An alternating voltage source given by \(e=140\sin(314.2\,t)\) is connected across a pure resistor of 50 \(\Omega\). Determine :(i) Frequency of the source.(ii) RMS value of emf.

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