Find the value of Leq in the given network.. Find V1, V2 and...

Find the value of $L_{eq}$ in the given network.. Find $V_1$, $V_2$ and $V_3$ in the circuit shown. Assume that all the capacitors had zero charge on them before the circuit was energized. All voltages are dc voltages. Determine the phase angle that $i_1(t)$ lags $v_1(t)$ and $v_1(t)$ leads $i_2(t)$, where $v_1(t) = 37 \sin(377t - 55^\circ) V$ $i_1(t) = -25 \cos(377t + 15^\circ) A$ $i_2(t) = -35 \sin(377t + 32^\circ) A$

Transcript

00:01 So here in the first part of this question, we are given the value of the operating voltage which is equals to 300 kilowatt.

00:12 We are also given the value of the line reactance and the value of the line reactance from here is equals to 200 ohm per phase and we are also given the per phase operating voltage that is equals to or we here is representing the operating voltage which is equals to 300 kilowatt per divided by under root 3 that from here become equals to 173 .2 kilowatt.

00:45 Now as we know that that the power per phase is given by the formula that is equals to p is equals to v multiplied by the i.

00:54 We can write it as v multiplied by the i.

00:57 I from here is equals to v divided by under root of line reactance that is j x.

01:05 So v multiplied by the v divided by the j x.

01:08 So this become equals to v square divided by the j x plugging into the value.

01:12 We are having the value of v which is equals to 173 .2 k raised to the power 2 which is divided by j multiplied by the 200.

01:22 So solving this term we get the value of the power per phase that from here is equals to 150 megawatt to the angle of minus 90 degree and we can say that the total power which is delivered by the line is equals to 3 multiplied by the 150 megawatt that from here is equals to 450 megawatt.

01:47 Hence this is the answer to the part a of this question.

01:51 Now in the part b of the question we are given that that the power delivered by the power delivered by the line is given which is equals to 100 megawatt.

02:08 We are also given the value of the reactance of the line.

02:14 So the reactance of the line is given that is equals to j multiplied by the 200 ohm.

02:20 We are that from here is equals to 200 to the angle of the 90 degree.

02:24 So v square of r from here become equals to 200 multiplied by the 100 of m to the angle of 90 degree.

02:31 So solving this term we get the value of vr which become equals to 141 .42 kilowatt...

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