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=√()2+R12 =√()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===√2A Phase of current in upper branch,
cos‌ϕ1=== ⇒ϕ1=45° Thus, in upper branch, current leads voltage by 45° as capacitor ispresent.
Current flowing in lower branch
i2 is
i2===2√2A Phase of current in lower branch is
cos‌ϕ2== ⇒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.