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