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]

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?

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]

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.

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]

We can combine the parallel current sources I_1 and I_2:

I = I_1 + I_2 = 10 mA + 40 mA = 50 mA ✔[/size][/color]

By the current divider rule,

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

[strike]There are two mistakes in this line, yet you end up with 10 mA and that's the right answer[/strike]https://www.physicsforums.com/images/icons/icon4.gif
Oops, my mistake. What you are using is not what I think of as current divider rule, but it's okay.

 

[JJBladester]

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.

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:

$$I_{R1}=\frac{(20\Omega )(.05 A)}{(25 \Omega )}=.04A$$

$$V_{R1}=\left (I_{R1} \right )\left (R_1 \right )=\left ( .04A \right )\left (25\Omega \right )=1V$$

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