MHT CET · Physics · Alternating Current
In LCR series circuit, an alternating e.m.f. 'e' and current ' \(i\) ' are given by equations \(\mathrm{e}=160 \sin (100 \mathrm{t})\) Volt and \(\mathrm{i}=250 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}\).
The average power dissipated in the circuit is
- A 2.5 W
- B 4.0 W
- C 10 W
- D 100 W
Answer & Solution
Correct Answer
(C) 10 W
Step-by-step Solution
Detailed explanation
\(e=1.60 \sin (100 t)\) volt and
\(\mathrm{i}=250 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \quad \mathrm{mA}\)
Comparing given equations with the standard forms, \(e=e_0 \sin \omega t\) and \(i=i_0 \sin (\omega t+\phi)\) we get,
\(\begin{aligned}
& \mathrm{e}_0=160 \mathrm{~V}, \mathrm{I}_0=250 \mathrm{~mA} \\
& \therefore \quad \text { Power }=\frac{\mathrm{e}_0}{\sqrt{2}} \cdot \frac{\mathrm{I}_0}{\sqrt{2}} \cos \phi \\
&=\frac{160 \times 250 \times 10^{-3}}{2} \times \cos \left(\frac{\pi}{3}\right) \\
&=\frac{40}{2} \times \frac{1}{2} \\
&=10 \mathrm{~W}
\end{aligned}\)
\(\mathrm{i}=250 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \quad \mathrm{mA}\)
Comparing given equations with the standard forms, \(e=e_0 \sin \omega t\) and \(i=i_0 \sin (\omega t+\phi)\) we get,
\(\begin{aligned}
& \mathrm{e}_0=160 \mathrm{~V}, \mathrm{I}_0=250 \mathrm{~mA} \\
& \therefore \quad \text { Power }=\frac{\mathrm{e}_0}{\sqrt{2}} \cdot \frac{\mathrm{I}_0}{\sqrt{2}} \cos \phi \\
&=\frac{160 \times 250 \times 10^{-3}}{2} \times \cos \left(\frac{\pi}{3}\right) \\
&=\frac{40}{2} \times \frac{1}{2} \\
&=10 \mathrm{~W}
\end{aligned}\)
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