KCET · Physics · Alternating Current
In an \(\mathrm{AC}\) circuit, \(V\) and \(I\) are given by \(V=150 \sin (150 t) \) volt and \(I=150 \sin \left(150 t+\frac{\pi}{3}\right)\) amp. The power dissipated in the circuit is
- A \(106 \mathrm{~W}\)
- B \(150 \mathrm{~W}\)
- C \(5625 \mathrm{~W}\)
- D zero
Answer & Solution
Correct Answer
(C) \(5625 \mathrm{~W}\)
Step-by-step Solution
Detailed explanation
Given \(V=150 \sin (150 t)\) volt
\(I=150 \sin (150 t+\pi / 3) \mathrm{amp}\)
\(I_{0}=150 \mathrm{amp}\)
\(V_{0}=150\) volt
\(P=\frac{1}{2} V_{0} I_{0} \cos \phi\)
\(P=0.5 \times 150 \times 150 \times \cos 60^{\circ}\)
\(P=5625 W\)
\(I=150 \sin (150 t+\pi / 3) \mathrm{amp}\)
\(I_{0}=150 \mathrm{amp}\)
\(V_{0}=150\) volt
\(P=\frac{1}{2} V_{0} I_{0} \cos \phi\)
\(P=0.5 \times 150 \times 150 \times \cos 60^{\circ}\)
\(P=5625 W\)
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