Alternating Current Part 1

Given, angular frequency, ω=100‌rad∕s
Capacitance of capacitor, C=100μF=100×10−6F
Inductance of inductor coil, L = 0.5 H
Resistance in upper branch, R1=100Ω
Resistance in lower branch, R2=50Ω,
In the given circuit consider current in upper branch be i1 and current flowing in lower branch be i2. The net current flowing in circuit will be I.
Impedance of upper branch can be calculated as
Z1=√XC2+R12=√(

1

ωC

)2+R12
=√(

1

100×100×10−6

)2+1002
=√1002+1002=100√2Ω
Impedance of lower branch can be calculated as
Z2=√XL2+R22
=√(ωL)2+R22
=√(100×0.5)2+502=√502+502
=50√2Ω
Current flowing in upper branch,
i1=

V

Z1

=

200

100√2

=√2A
Phase of current in upper branch,
cos‌ϕ1=

R1

Z1

=

100

100√2

=

1

√2

⇒ϕ1=45°
Thus, in upper branch, current leads voltage by 45° as capacitor ispresent.
Current flowing in lower branch i2 is
i2=

V

Z2

=

200

50√2

=2√2A
Phase of current in lower branch is
cos‌ϕ2=

R2

Z2

=

50

50√2

⇒ϕ2=45°
Thus, in lower branch current lags voltage by 45° as inductor ispresent.
Thus, the net current, I=√i12+i22
I=√(√2)2+(2√2)2
I=√10=3.16A
Thus, no option in the given question is correct.
If I=i1+i2 is taken then I=√2+2√2=4.24A is obtained which is incorrect method of solution.