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NCERT Class XII Chapter
Alternating Current
Questions With Solutions

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Question : 19 of 26
Marks: +1, -0
Suppose the circuit in Q. 18 has a resistance of 15 Ω. Obtain the average power transferred to each element of the circuit and the total power absorbed.
Solution:  
If the circuit has a resistance of 15 Ω, now it is LCR series resonant circuit.
Now the impedance,
Z = R2+(XLXC)2\sqrt{R^2+(X_L-X_C)^2}
Z = 152+(27.93)2\sqrt{15^2+(27.93)^2} = 31.7 Ω
Virtual current, IvI_v = EvZ\frac{E_v}{Z} = 23031.7\frac{230}{31.7} = 7.26 A
Average power transferred to ‘L’,
PLP_L = IvEvI_v E_v cos π/2 = 0
Average power transferred to ‘C’,
PCP_C = EvIvE_vI_v cos π/2 = 0
Average power transferred to ‘R’,
PRP_R = VRIvV_RI_v cos 0°
PRP_R = (IvR)Iv(I_v R) I_v = Iv2I_v^2 R = (7.26)2 × 15 = 791 W
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