[2pts] Consider the network above, with six points labeled 1...

[2pts] Consider the network above, with six points labeled 1 through 6. $R_1 = 20M\Omega$, $R_2 = 80M\Omega$, $V_0 = 738V$, and $I_0 = -3.6\mu A$. The current through $V_0$ is $-4.5\mu A$ (arrow to the left). The current from point 5 to point 2 (by the most direct path) is $8.1\mu A$. a. What is the power absorbed by $R_2$? Answer: b. What is the power delivered by $V_0$? Answer:

Transcript

00:01 Hello student, in this question given as a circuit.

00:02 Now, first of all in the first of first part of the question, we have to calculate the nodal potential node at the node potential v.

00:12 We will apply the kirchhoff current law.

00:15 So i1 plus i2 plus i3 will be equal to the 0.

00:22 So i1 is equal to that will be v plus 20 upon 2.

00:30 I2 is equal to i2 will be equal to the v minus 30 upon 2 plus i3 is equal to v minus 40 upon r out is equal to 0.

00:49 Now in the first part of the question given as a if r out is equal to 3 kilovolt, we will substitute their values of r out and after calculation we will get the potential at node will be equal to the 110 upon 8 volt and we will calculate the i3 current.

01:12 So this will be equal to the v minus 40 upon 3 that is minus 70 upon 8 milliampere.

01:26 Negative sign shows that the current flowing in the anti -clockwise direction.

01:32 Now, we will calculate the power delivered to the across the r out.

01:42 So power delivered across the r out will be equal to the i3 square into r.

01:52 I3 is the 70 by 8 whole square into r is 3 kilo ohm.

02:01 So the power will be equal to the across the r out will be equal to 229 .6875 milliwatt.

02:15 In the second part of the question, we have to calculate the hevenin voltage.

02:24 First step is we remove the r out and apply the kvl in first loop.

02:32 So this 30 volt plus 2 i plus 2 i plus 20 is equal to 0.

02:42 So i will be equal to the minus 12 .5...

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