Finding resistor's voltage with current and voltage sources present

JJBladester

What is the voltage across R2? You should convert the 4 V source and resistor into a current source as your first step.



The first thing I did is create a second current source out of the 4 V voltage source and the series resistor R2. I called it I_2.

I_2 = E/R_2 = 4V/100Ω = 40 mA

This gives us:



We can combine the parallel current sources I_1 and I_2:

I = I_1 + I_2 = 10 mA + 40 mA = 50 mA



By the current divider rule,

I_R2 = (R_T*I)/R2 = (20Ω*50mA)/100Ω = 10 mA

V_R2 = I_R2*R2 = 10mA * 100Ω = 1 V

However, when I run a simulation of the original circuit using MultiSim, I get 3 V through R2. Where am I going wrong?

SammyS

If you get 3 V across R₂ in the original circuit, that means that there voltage dropped across R₁ in the original circuit is 1 V, which makes sense.

The difficulty appears to be that you are assuming the resistor, R₂ in the original circuit has the same voltage across it as the resistor, R₂, in the last two circuits.

JJBladester

I'm stuck. I think I understand what you mean that R2 in the new (simplified) circuit cannot be treated the same as R2 in the original circuit. Am I correct up to the point where I found a combined 50 mA current source acting on two parallel resistors R1 and R2?

If so, how do I get back to finding the voltage for R2 in the original circuit.

SammyS

Find the voltage across R₁ in the simplified circuit. It hasn't been changed. Then use that to find the voltage across R₂ in the original circuit.

NascentOxygen

There are two mistakes in this line, yet you end up with 10 mA and that's the right answer
Oops, my mistake. What you are using is not what I think of as current divider rule, but it's okay.

JJBladester

So I can't use the "new" R₂ even though it is the same R₂ as before?

Anyway, using your advice to find the voltage across R₁ in the simplified circuit, I've found the voltage across R₂ using Kirchhoff's Voltage Law on the original circuit:



Using the current divider rule:

[tex]I_{R1}=\frac{(20\Omega )(.05 A)}{(25 \Omega )}=.04A[/tex]

[tex]V_{R1}=\left (I_{R1} \right )\left (R_1 \right )=\left ( .04A \right )\left (25\Omega \right )=1V[/tex]

E - V_R2 - V_R1 = 0
V_R2 = E - V_R1 = 4V - 1V = 3V