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Question

Figure shown parallel combination of L,C and R connected to an A.C source supplying current I=14A.IN the circuit R1=8Ω,R2=6Ω,XL=6Ω and XC=8Ω.find the power dissipated in the circuit
1068356_719d87f08e8a4fa0ac27e2086c228537.png

A
686W
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B
1372W
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C
920W
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D
1400W
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Solution

The correct option is B 1372W
Ref. image
R1=8Ω R2=6Ω
XL=6Ω XC=8Ω
ZL=RL+XL
=8Ω+6Ω
ZL=14Ω
ZC=RZ+XC
6Ω+8Ω
ZC=14Ω
12+1ZL+1ZC
12=114+114
Z=7Ω
Power =I2Z
=14×14×7
P=1372w

1431354_1068356_ans_34bbc581d32147f586e359460e3a76ab.png

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