MHT CET · Physics · Alternating Current
In an AC circuit \(\mathrm{E}=200 \sin (50 \mathrm{t})\) volt and \(I=100 \sin \left(50 t+\frac{\pi}{3}\right) \mathrm{mA}\). The power dissipated in the circuit is \(\binom{\sin 30^{\circ}=\cos 60^{\circ}=0.5}{\sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3} / 2}\)
- A 20 watt
- B 15 watt
- C 10 watt
- D 5 watt
Answer & Solution
Correct Answer
(D) 5 watt
Step-by-step Solution
Detailed explanation
\(e=200 \sin (50 t) \text { and } I=100 \sin \left(50 t+\frac{\pi}{3}\right)\)
Comparing these equations with the standard forms \(\mathrm{e}=\mathrm{e}_0 \sin \omega \mathrm{t}\) and \(\mathrm{I}=\mathrm{I}_0 \sin \omega \mathrm{t}\), we get, \(\mathrm{e}_0=200 \mathrm{~V}, \mathrm{I}_0=100 \times 10^{-3} \mathrm{~A}, \phi=\) phase difference \(=60^{\circ}\)
\(\therefore P=e_{r m s} I_{r m s} \cos \phi =\frac{e_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \times \cos 60^{\circ}\)
\(=\frac{200}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \frac{1}{2}=5 \mathrm{~W}\)
Comparing these equations with the standard forms \(\mathrm{e}=\mathrm{e}_0 \sin \omega \mathrm{t}\) and \(\mathrm{I}=\mathrm{I}_0 \sin \omega \mathrm{t}\), we get, \(\mathrm{e}_0=200 \mathrm{~V}, \mathrm{I}_0=100 \times 10^{-3} \mathrm{~A}, \phi=\) phase difference \(=60^{\circ}\)
\(\therefore P=e_{r m s} I_{r m s} \cos \phi =\frac{e_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \times \cos 60^{\circ}\)
\(=\frac{200}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \frac{1}{2}=5 \mathrm{~W}\)
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