In fig27 18 1 100v 2 300v r1 400 r2 200 r3 500 and both...

In Fig.27-18, 1 1.00V, 2 3.00V, R1 4.00 ,R2 2.00 ,R3 5.00 , and both batteries are ideal.What isthe rate at which energy is dissipated in (a) R1,(b) R2,and (c) R3? What is the power of (d) battery 1 and (e) battery 2? Is energy being absorbed or provided in (f) battery 1 and (g) battery 2?

Transcript

00:01 In this problem, we have been given the circuit and even an ie2 are the potential difference across the two cells whose values are given here and the resistor values are also indicated.

00:12 So we need to determine first the power across the resistor r1.

00:18 So for power, we're going to use the equation either v square by r or v into i or either i square into r.

00:28 So for that here let's figure out the current that's flowing in each part first.

00:34 So let's say from e1, ivan current is emitted and from e2, i2 amount of current is emitted here.

00:41 So through this r3, i1 plus i2 current will flow.

00:45 So we apply kirchav's loop law here to get first the current.

00:51 So in the loop here, there's first one we apply kirchav's voltage law.

00:56 So according to that, the equation comes.

00:58 Comes out to be 0 plus 1 .45 minus there will be potential drop across this r1 that will be obtained using omslaw.

01:07 So i1 into the resistance 4 .37.

01:11 Again there will be potential drop across this r3 so that will be minus 4 .62 times i1 plus i2 and that's equal to 0.

01:21 So here this is equation 1.

01:23 Similarly we consider another loop and we apply kirchow's voltage law to the second loop.

01:29 And the equation comes out to be 1 .45 minus i2 into 1 .52 minus 4 .62 into i1 plus i2 and this is again equal to 0 so this is equation 2 so let's just change this value of even that we have 3 .27 volt so here we simplify the equation so the first equation that we get here is 3 .27 minus 4 .37 and 4 .62 we just add them together to get 8 .99 i1 and minus 4 .62 i2 and that's equal to 0.

02:21 So this can once again be simplified as 8 .99 i1 minus 4 point.

02:28 This will go to the other side.

02:30 It will become plus so plus 4 .62 i2 minus 3 .27 equals to 0 .7.

02:36 So this is equation 1.

02:38 We also simplify equation 2.

02:40 So that will be minus 1 .45.

02:45 And once again we add minus 1 .52 and minus 4 .62 into i2.

02:51 That gets us 6 .14 i2.

02:57 And this will also be negative.

03:00 So also we have 4 .62 i1...