S2E3: Using KVL, KCL, and VI constraints

In the network shown you are given that V, A, , and .

Wow, this one was a little tricky. I went to the forums and found an answer by UXO:

Short answer: apply KVL to the left-hand loop:
-V + i1R1 + i2R2 = 0
Then use your observation that i2 = i1 + I
=> -V + i1R1 + i1R2 + I*R2 = 0
i1 (R1 + R2) = V - I*R2
i1 = (V-(I R2))/(R1 + R2) = (2-3*5)/(4+5) = -1.44
Then just plug values back in and determine the rest of the unknowns.

So to recap, we KVL the left-hand loop:

-V + (i1*R1) + (i2*R2) = 0

Then we do a KCL on the node above R2:

-i1 + i2 - I = 0

Now we solve the KCL equation in terms of i2:

i2 = i1 + I

Then we substitute that into the KVL equation:

-V + (i1*R1) + ((i1 + I)*R2) = 0

Now we distribute R2 over the replaced i2:

-V + (i1*R1) + (i1*R2) + (I*R2) = 0

Then we solve this equation in terms of i1:

(i1*R1) + (i1*R2) + (I*R2) = V

(i1*R1) + (i1*R2) = V - (I * R2)

i1 * (R1 + R2) = V - (I * R2)

i1 = (V - (I * R2)) / (R1 + R2)

And then we plug in the values:

i1 = (2.0 - (3.0 * 5.0)) / (4.0 + 5.0) = (2.0 - (15.0)) / (9.0) = -13.0 / 9.0 = -1.44 Amps

With that in hand, we can determine the other currents:

-i1 + i2 - I = 0

i2 = I + i1

i2 = 3.0 - 1.44 = 1.56 Amps

And with those in hand, we can determine the voltages:

v1 = i1 * R1

v1 = -1.44 * 4.0 = -5.76 Volts

v2 = i2 * R2

v2 = 1.56 * 5.0 = 7.8 Volts

What is the voltage (in Volts) across the resistor with resistance ?

7.8

What is the power (in Watts) dissipated by the resistor with resistance ?

P = IV

P = i2 * v2

P = 1.56 * -5.76 = 12.2 Watts

What is the current (in Amperes) through the resistor with resistance ?

-1.44

What is the power (in Watts) dissipated by the resistor with resistance ? 

P = IV

P = i1 * v1

P = -1.44 * -5.76 = 8.3 Watts

What is the power (in Watts) supplied by the voltage source?

P = IV

P = -1.44 * 2.0 = -2.88 Watts

What is the power (in Watts) supplied by the current source?

P1 + P2 = Pr1 + Pr2 

P2 = Pr1 + Pr2 - P1

P2 = 8.3 + 12.2 - (-2.88) = 23.4 Watts

You should observe that the sum of the power supplied by the sources is the sum of the power dissipated by the resistors. If this is not true you have done something wrong.