Using current division to find voltage

Question

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\$.

Here is my 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.

Answer 1

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

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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 $$V_L = (i_1 - i_2)R$$

Answer 2

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.