Apply the loop rule to loop akledcba going clockwise in the...

Name: apply the loop rule to loop akledcba going clockwise in the figure below using the direction of the currents as shown on the figure write your expression using the following symbols for the 22103
Uploaded: 2023-06-30T10:56:41-08:00
Duration: 6 min 56 s
Channel: Sri K
Description: apply the loop rule to loop akledcba going clockwise in the figure below using the direction of the currents as shown on the figure write your expression using the following symbols for the 22103

Apply the loop rule to loop akledcba (going clockwise) in the figure below, using the direction of the currents as shown on the figure. Write your expression using the following symbols for the emf, currents and resistances: E1 for $E_1$, E2 for $E_2$, E3 for $E_3$, E4 for $E_4$, I1 for $I_1$, I2 for $I_2$, I3 for $I_3$, R1 for $R_1$, R2 for $R_2$, R3 for $R_3$, R5 for $R_5$, r1 for $r_1$, r2 for $r_2$, r3 for $r_3$, and r4 for $r_4$. $E_1 = 24.0$ V $R_1$ 5.0 $\Omega$ $E_2 = 48.0$ V $I_1$ $I_2$ $R_3$ 78 $\Omega$ $I_3$ $E_3 = 6.0$ V $r_5$ 0.05 $\Omega$ $r_1$ 0.10 $\Omega$ $r_2$ 0.50 $\Omega$ $R_2$ 40 $\Omega$ $R_5$ 20 $\Omega$ $r_4$ 0.20 $\Omega$ $E_4 = 36.0$ V

Transcript

00:01 So here we have to solve the given question using the kirchhoff's law where we are given the value of let's say e1 that is equals to 70 .8 watt.

00:11 We are given the value of e2 that is equals to 60 .2 watt.

00:15 We are having the value of e3 that is equals to 79 .2 watt.

00:20 So here we are given a circuit which is consisting of two adjacent rectangular loop.

00:25 Let's we can make the circuit from here.

00:27 So the system is somewhat like this.

00:30 Here is the first battery is connected that from here is equals to 60 .2 volt.

00:36 Here the another battery is connected that from here is equals to 70 .8 volt and then another battery from here is connected in this way e3 that is equals to 79 .2 volt.

00:50 So here the resistance let's say r1 resistance is here.

00:54 Here the r2 resistance is connected and from here let's say here is another r3 resistance.

01:02 So this is the whole system here.

01:05 We are having e1 this is e1 this is e2 this is e3 resistance r1 is here which is equals to 2 kilo ohm resistance r2 is equals to 3 kilo ohm and resistance r4 is equals to 4 kilo ohm.

01:19 So the current from here is moving in this way.

01:22 Let's say this is i1 this from here is i2 and this from here i3.

01:28 So current is moving from here in this way.

01:31 So from here what we can do that we are considering let's say this is a system is a loop a side b c d e and this point is f.

01:43 Now we are considering for the loop a b c f and a.

01:48 So we can apply the kirchhoff's law here.

01:51 So after applying the kirchhoff's equation will be like e1 minus e2 minus i2 multiplied by the r2 minus i1 r1 that is equals to zero.

02:03 So what we can do from here is we can separate the term of e and i from here.

02:08 So i1 multiplied by the r1 plus i2 multiplied by the r2 is equals to e1 minus e2.

02:15 Plugging into the value e1 from here is equals to 70 .8 minus e2 that is 60 .2 volt.

02:23 This from here is i1 r1 plus i2 r2.

02:27 So the value that is i1 r1 plus i2 r2 from here become equals to 10 .6 volt.

02:37 Let's say this is the equation number one.

02:40 Now when we apply the kirchhoff's law in the loop e c d e f c.

02:46 So we get the equation somewhat like this that is e2 multiplied by the i3 r3 minus e3 plus i2 r2 that is equals to zero.

02:58 Further simplifying the equation that is i2 r2 plus i3 r3 that is equals to e3 minus e2 that from here again become equals to that from here again become equals to 79 .2 minus 60 .2.

03:16 So this equation from here become equals to i2 r2 plus i3 r3 that is equals to 19 volt...

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