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Alternating Current

Question

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An inductor 20 mH, a capacitor 100 µF 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

0.79 W

0.43

1.13

2.74 W

Solution

0.79 W

Power dissipated in an LCR series connected to an a.c source of emf E

$P = E_{r m s} i_{r m s} \cos ϕ = \frac{{E^{2}}_{r m s} R}{Z^{2}} = \frac{E_{r m s}^{2} R}{R^{2} + {(ω L - \frac{1}{C ω})}^{2}} = \frac{{(\frac{10}{\sqrt{2}})}^{2} x 50}{(50)^{2} + {(314 x 20 x 10^{- 3} - \frac{1}{314 x 100 x 10^{- 6}})}^{2}} s o l v i n g w e g e t, P = 0.79 W$

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Alternating Current

An inductor 20 mH, a capacitor 100 µF 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

0.79 W

0.43

1.13

2.74 W

Some More Questions From Alternating Current Chapter

Alternating current can not be measured by D.C. ammeter because

In an LCR series a.c. circuit, the voltage across each of the components, L, C and R is 50 V. The voltage across the LC combination will be

If we change the value of R, then

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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For Daily Free Study Material Join wiredfaculty

Alternating Current

An inductor 20 mH, a capacitor 100 µF 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 0.79 W 0.43 1.13 2.74 W

Some More Questions From Alternating Current Chapter

Alternating current can not be measured by D.C. ammeter because

In an LCR series a.c. circuit, the voltage across each of the components, L, C and R is 50 V. The voltage across the LC combination will be

If we change the value of R, then

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

An inductor 20 mH, a capacitor 100 µF 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

0.79 W

0.43

1.13

2.74 W