Question

An AC source is \(120\) V - \(60\) Hz. The value of voltage after \(\frac{1}{720}\) s from start will be \(\left(\sin 30^{\circ }=\frac{1}{2}\right)\)

• A. \(20.2V\)
• B. \(42.4V\)
• C. \(84.8V\)
• D. \(106.9V\)

Hint:

The quoted 120 V is the rms value, so the peak is 120 times root 2, then evaluate V equal to V0 sin omega t.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
The rated value of an AC source is the rms value. Peak voltage \(V_0=\sqrt{2}\,V_{rms}=120\sqrt{2}\approx169.7\) V.

Step 2: Phase angle
\(\omega t=2\pi\times60\times\dfrac{1}{720}=\dfrac{\pi}{6}=30^{\circ}\).

Step 3: Instantaneous value
\(V=V_0\sin\omega t=169.7\times\sin30^{\circ}=169.7\times\dfrac{1}{2}\).

Step 4: Result
\(V\approx84.8\) V, option (C).

Final Answer:
The voltage after 1 by 720 s is about 84.8 V. \[ \boxed{\text{(C) }84.8\,V} \]