Recognitions:
Gold Member

## Finding resistor's voltage with current and voltage sources present

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?
 PhysOrg.com science news on PhysOrg.com >> King Richard III found in 'untidy lozenge-shaped grave'>> Google Drive sports new view and scan enhancements>> Researcher admits mistakes in stem cell study

Mentor
 Quote by 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?
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.
 Recognitions: Gold Member 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.

Mentor

## Finding resistor's voltage with current and voltage sources present

 Quote by 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.
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.

Recognitions:
Homework Help
 Quote by JJBladester 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
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.

Recognitions:
Gold Member
 Quote by 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.
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 - VR2 - VR1 = 0
VR2 = E - VR1 = 4V - 1V = 3V

 Tags current source, resistor, simulation, voltage source