8 the resistances in the figure below are all 60 omega and...

8. The resistances in the figure below, are all $6.0 \Omega$ and the batteries are ideal $24V$ batteries. (a) When switch S in the figure a is closed, what is the change in the electric potential $V_1$ across resistor 1, or does $V_1$ remain the same? (b) When switch S in the figure b is closed, what is the change in $V_1$ across resistor 1, or does $V_1$ remain the same? S S $R_3$ $R_1$ $R_1$ $R_2$ (a) (b) Problem-8

Transcript

00:01 So in order to solve these problems, we will apply kitchf's loop rule to each of the parts before the switch is closed and after the switch is closed.

00:13 So when the switch was open, we have the loop rule to be like this.

00:20 So let me just draw the loop.

00:25 So my first loop is like this and my second loop.

00:29 Let's say like this.

00:30 Now i'll apply the loop rule to each of these two loops and solve for the voltage across r1 for that.

00:42 Now the loop rule for loop 1 will look like.

00:48 So let's say my current is going like this.

00:52 So this is i and therefore this part is i1 and this is i2.

01:00 Now when the switch is open, there is no current flowing through the second resistor, which means that i2 will be 0 when the switch is open, and therefore we just have to consider the first loop.

01:17 So we are going from negative to positive, so plus emf, and then we are going along the direction of current at r1, so minus i1r1, and that is equal to 0.

01:33 Which means that i1 will be emf over r1, and therefore the voltage across r1, which is v1, will be i1 times r1, therefore this will be equal to e.

01:48 So now the voltage is just equal to the emf of the battery that is given to be 12 volts.

01:55 And this is the case before the switch is closed.

02:01 Now, when the switch is closed, so let me draw this with the green sign.

02:13 So now the switch is closed.

02:15 So for that, the loop rule for the first loop will look like this.

02:24 And there is no need to write the equation for the second loop.

02:30 Only the result from the first loop is enough because, you see from this equation we can find i1 in terms of r1 and therefore this will mean that v1 will be equal to i1 times r1 and hence this will be equal to the emf which is equal to 12 volts.

02:57 And therefore we see that before and after the switch is open, the voltage remains the same.

03:05 It's not changed.

03:07 It has the same value.

03:08 So the answer is v1 is same.

03:16 Now coming to the second part, we will follow the same method.

03:21 So let me just draw my loops.

03:26 I'm choosing the same orientations of the loop as i chose over here.

03:31 So now let's write the, okay, before that, we need to mark the currents.

03:38 So this is i3, this is i1.

03:41 Based on their resistors and denoting the currents and this is i2...