Choose a voltage V <2.5 V

1.Choose a voltage v < 2.5 V and values for the resistors R1, R2, R3, and R4 in
the circuit of Fig. 3.90 so that i1 =1 A, i2 =1.2 A, i3 =8 A, and i4 = 3.1 A.

KCL/KVL/OHms Law

The Attempt at a Solution

If Is = I1 + I2 + I3 + I4,

Then, Is = $\frac{v}{Req}$

Req = $\frac{1}{R1}$ + $\frac{1}{R2}$ + $\frac{1}{R3}$ + $\frac{1}{R4}$.

After this, I get stuck...

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• 3.90.png
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You're over thinking the problem. For parallel branches, the voltage is the same to all branches of the node. For simplicity choose 1V and for R1 choose 1 ohm so that I1 = 1A. Now you have the voltage and current for each branch so just solve for resistance via ohm's law.

You're over thinking the problem. For parallel branches, the voltage is the same to all branches of the node. For simplicity choose 1V and for R1 choose 1 ohm so that I1 = 1A. Now you have the voltage and current for each branch so just solve for resistance via ohm's law.

So because we can choose any v that is less than 2.5V, you picked 1V correct?

Taking V as 1A and using the given values for the currents, I solved for R1, R2, R3, and R4:

R1 = 1V/1A = 1Ω

R2 = 1V/1.2A = 0.83Ω

R3 = 1V/8A = 0.125Ω

R4 = 1V/3.1A = 0.323Ω

Is this correct?

Looks good. And if for some reason you need the voltage to be higher than you can just scale the resistance proportionately to maintain the same current.