An inductor of reactance 1Ω and a resistor of 2Ω are connected in series to the terminals of a 6V(rms) a.c. source. The power dissipated in the circuit is
A
8W
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B
12W
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C
14.4W
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D
18W
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Solution
The correct option is C14.4W
Here, XL=1Ω,R=2Ω,Vrms=6V Impedence of the circuit, Z=√X2L+R2=√(1)2+(2)2=√5Ω Irms=VrmsZ=6√5A Power dissipated, P=VrmsIrmscosϕ=VrmsIrmsRZ =6×6√5×2√5=725=14.4W