I'm a beginner in electrical engineering. I am having trouble with this circuit analysis problem.

It asks to simplify the circuit into an expression for \$V_L\$ in terms of \$R, R_L,\$ and \$i_S\$.

Initial Problem

Here is my work:

Handwritten Work

Using Kirchhoff's laws and Ohm's law, I get \$i_S = i_1 + i_2\$ and \$V_L = (Ri_1) + (Ri_2)\$.

Taking the GCF of \$R\$ gets \$V_L = R(i_1 + i_2)\$ therefore \$V_L = Ri_S\$.

Am I doing this correctly? Is there a way to simplify the resistors \$R\$ and \$R_L\$ and find the voltage \$V_L\$? Any help to point me in the right direction is appreciated.

  • \$\begingroup\$ But V_L is a voltage drop across R_L alone. \$\endgroup\$
    – G36
    Sep 9, 2021 at 18:06

2 Answers 2


Your equation for \$i_S\$ is correct but the one for \$V_L\$ is not -- you've applied KVL incorrectly, and the \$i_2\$ term should be negative.1

In any case, the problem is easily solved using the current divider rule, which states that the current in one of two branches is the total current entering the branches (\$i_S\$ here) times the resistance of the other branch divided by the sum of the resistances of the two branches. Therefore

$$i_2 = i_S \frac{R}{R + R + R_L} = i_S \frac{R}{2R + R_L}$$

We also know by inspection that $$V_L = i_2 R_L$$

Therefore $$V_L = i_S \frac{R R_L}{2R + R_L}$$

1KVL states that the voltage at node A is $$V_A = i_1 R = i_2(R + R_L) = i_2 R + V_L$$ which simplifies to $$V_L = (i_1 - i_2)R$$


It asks to simplify the circuit into an expression for \$V_L\$

\$R_{(\text{as seen by } i_s)} = R\cdot\dfrac{R+R_L}{2R+R_L}\$ (parallel resistor rule)

Voltage across source \$i_s\$ is therefore \$i_s\cdot R\cdot\dfrac{R+R_L}{2R+R_L}\$.

Using the voltage divider rule, \$V_L = i_s\cdot \dfrac{R+R_L}{2R+R_L}\cdot \dfrac{R_L}{R+R_L} = i_s\cdot \dfrac{R\cdot R_L}{2R+R_L}\$.

Am I doing this correctly?

It seems you aren't.


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