Question #142418
If an R = 1-kΩ resistor, a C = 1-µF capacitor, and an L = 0.2-H inductor are connected in
series with a V = 150 sin (377t) volts source, what is the maximum current delivered by the source?
1
Expert's answer
2020-11-09T14:18:53-0500

Solution: Based on the datas given in the question a circuit diagram can be constructed as shown in the below figure.





From this we can calculate capacitive reactance = Xc=1/Cω=1/(106×377)=2652.52ΩX_c = 1/C\omega = 1/(10^{-6} \times377) = 2652.52\varOmega


The inductive reactance =

XL=Lω=0.2×377=75.4Ω.X_L = L\omega = 0.2\times377 = 75.4 \varOmega.


The rms value of input voltage =150/2=106.07V= 150/\sqrt{2} = 106.07V


Hence an equivalent circuit in the frequency domain incorporating the sign of capacitive and inductive reactance is as shown below





Total impedance of the circuit = ZT=R+j(XLXc)=1000+j(752652.52)Z_T =R + j(X_L - X_c) = 1000 +j(75-2652.52)


ZT=1000+j2577.12  ΩZ_T = 1000 + j2577.12 \space\space\varOmega


Current through the circuit = V/ZT=106.07/(1000j2577.12)=0.01388+j0.03577V/Z_T = 106.07/(1000-j2577.12) = 0.01388 +j0.03577

There for current I=0.0383768.79°I = 0.03837\angle68.79\degree


We obsereve the rms value = 38.37mA38.37 mA


Maximum value of current = rmsvalue×2=38.37×2=54.26mArms value \times\sqrt{2} = 38.37\times\sqrt{2} = 54.26 m A


The current can be written in sinusoidal form as 54.26Sin(377t+68.79°)mA54.26Sin (377t+68.79\degree) m A


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