Answer to Question #142418 in Electrical Engineering for Jisha

Question

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?

Answer 1

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 = $X_c = 1/C\omega = 1/(10^{-6} \times377) = 2652.52\varOmega$

The inductive reactance =

$X_L = L\omega = 0.2\times377 = 75.4 \varOmega.$

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

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 = $Z_T =R + j(X_L - X_c) = 1000 +j(75-2652.52)$

$Z_T = 1000 + j2577.12 \space\space\varOmega$

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

There for current $I = 0.03837\angle68.79\degree$

We obsereve the rms value = $38.37 mA$

Maximum value of current = $rms value \times\sqrt{2} = 38.37\times\sqrt{2} = 54.26 m A$

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