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Question: An inductor of reactance \(1 \Omega\) and a resistor of \(2 \Omega\) are connected in series to the...

An inductor of reactance 1Ω1 \Omega and a resistor of 2Ω2 \Omega are connected in series to the terminals of a 6 V (rms) ac source. The power dissipated in the circuit is

A

8 W

B

12 W

C

14.4 W

D

18 W

Answer

14.4 W

Explanation

Solution

: Here, XL=1Ω,R=2Ω, Vrms=6 V\mathrm { X } _ { \mathrm { L } } = 1 \Omega , \mathrm { R } = 2 \Omega , \mathrm {~V} _ { \mathrm { rms } } = 6 \mathrm {~V}

Impedance of the circuit,

Z=XL2+R2=(1)2+(2)2=5Ω\mathrm { Z } = \sqrt { \mathrm { X } _ { \mathrm { L } } ^ { 2 } + \mathrm { R } ^ { 2 } } = \sqrt { ( 1 ) ^ { 2 } + ( 2 ) ^ { 2 } } = \sqrt { 5 } \Omega

Irms=VrmsZ=65 AI _ { \mathrm { rms } } = \frac { V _ { \mathrm { rms } } } { Z } = \frac { 6 } { \sqrt { 5 } } \mathrm {~A}

Power dissipated,

=6×65×25=725=14.4 W= 6 \times \frac { 6 } { \sqrt { 5 } } \times \frac { 2 } { \sqrt { 5 } } = \frac { 72 } { 5 } = 14.4 \mathrm {~W}