An inductor 20 mH , a capacitor 100 μF and a resistor 50 Ω are connected in series across a source of emf, V = 10 (314 t) . The power loss in the circ - Physics MCQ NEET (UG) 2018 2026 | ExamTest.live

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An inductor 20 mH, a capacitor 100 μ F and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin (314 t) . The power loss in the circuit is:

Solution & Step-by-step Explanation

1. From V = 10 sin (314 t), we have V_{ma x} = 10 V and ω = 314 rad/s .2. X_{L} = ω L = 314 \times 20 \times 1 0^{- 3} = 6.28 Ω .3. X_{C} = \frac{1}{ω C} = \frac{1}{314 \times 100 \times 1 0 ^{- 6}} \approx 31.85 Ω .4. Impedance Z = R^{2} + (X_{C} - X_{L})^{2} = 5 0^{2} + (31.85 - 6.28)^{2} = 2500 + 653.8 \approx 56.16 Ω .5. Power loss P = V_{r m s} I_{r m s} cos ϕ = \frac{V _{r m s}^{2} R}{Z ^{2}} . P = \frac{( 10/ 2 ) ^{2} \times 50}{( 56.16 ) ^{2}} = \frac{50 \times 50}{3154} \approx 0.79 W

Try it yourself before checking the explanation above.

An inductor 20 mH, a capacitor 100 μ F and a resistor 50 Ω are connected in series across a source of emf, V = 10 sin (314 t) . The power loss in the circuit is:

2.74 W

0.43 W

0.79 W

1.13 W

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