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Q.

For the circuit shown in the figure, find the peak current (in ampere) through the source. E = 50 sin (100t)

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answer is 5.

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Detailed Solution

XL=ωL=100(0.1)=10Ω  Z1=R2+XL2=102+102=102XC=1ωC=1100(103×10-6)=10ΩZ2=R2+XC2=102+102=102 Peak current through  R1,I1=50102=52APeak current through  R2,I2=50102=52APhase difference between I1 and I2 is  π2∴  Peak current through the source is I=I12+I22=522+52=52×2∴       I=5  A
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For the circuit shown in the figure, find the peak current (in ampere) through the source. E = 50 sin (100t)