*10.7 In the circuit of Fig. P10.7, VaN = 65∠47^∘ V (rms), VbN = 60∠15^∘ V (rms), VcN = 45∠-86^∘

Step 1: Convert the given voltages from phasor form to time-domain form using the formula V(t) =

*10.7 In the circuit of Fig. P10.7, VaN = 65∠47^∘ V (rms),...

*10.7 In the circuit of Fig. P10.7, $V_{aN} = 65\angle47^\circ$ V (rms), $V_{bN} = 60\angle15^\circ$ V (rms), $V_{cN} = 45\angle-86^\circ$ V (rms), and $f = 60$ Hz. Calculate $v_{ab}(t)$, $v_{bc}(t)$, and $v_{ca}(t)$.

Transcript

00:01 Here in part a we are asked to write the equation for loop a .e.

00:05 Dc, b, a.

00:06 And in this question, in part 2, we are asked to find the value of current i2.

00:11 So let's start to solve this problem.

00:14 Here, first we solve part a, means we write the loop equation for loop ae, b ,c, b, a.

00:21 Here, current flowing from right to left means here this terminal is positive and this terminal is negative.

00:28 In another words, we can say, due to resistance, a voltage drop takes place.

00:34 Means here, this terminal is the higher potential and this terminal is the lower potential.

00:39 And here, current flowing from higher potential to lower potential.

00:44 Similarly, in this resistance, current flowing from this terminal to this terminal, means here this terminal is higher potential and this terminal is the lower potential.

00:55 And in this resistance, same i2 current is flowing.

00:58 Current is flowing from left to right means here this terminal is positive and this terminal is negative means here this terminal is the higher potential and this terminal is the lower potential now next we apply kirchof voltage low in this loop and by applying kirchof voltage low in this loop we obtain this equation here we start to write the kvl equation from this resistance so here we moving from higher potential to lower potential means voltage drop takes place and across resistance i into our voltage drop takes place, means i2 into 2 .5.

01:39 And when we goes in this direction, then due to this battery, a voltage rise takes place because in the battery current is flowing from lower potential to higher potential, means positive 18.

01:55 Similarly here current leaves the negative terminal of this resistance, means negative i.

02:01 Into r i2 into 0 .5 here the current is leaving the negative terminal of this resistance means negative i1 into 6 and by kirchow voltage law we can say the algebra sum of voltage drop in a loop comes equal to 0 now from this we obtain 6 i1 plus 3 i2 equals to 18 and after dividing this equation by 3 we obtained 2 i1 plus i2 equals to 6 so here this is the kbl equation for this loop a dc b a suppose here this is equation number 1 now next for solving part b we write the kvl equation in this loop and in this loop we start to write the kvl equation from the point h...

Question

*10.7 In the circuit of Fig. P10.7, $V_{aN} = 65\angle47^\circ$ V (rms), $V_{bN} = 60\angle15^\circ$ V (rms), $V_{cN} = 45\angle-86^\circ$ V (rms), and $f = 60$ Hz. Calculate $v_{ab}(t)$, $v_{bc}(t)$, and $v_{ca}(t)$.

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