Register to reply

Finding resistor's voltage with current and voltage sources present

Share this thread:
JJBladester
#1
Jul16-12, 09:16 PM
PF Gold
JJBladester's Avatar
P: 287
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?
Phys.Org News Partner Science news on Phys.org
Final pieces to the circadian clock puzzle found
A spray-on light show on four wheels: Darkside Scientific
How an ancient vertebrate uses familiar tools to build a strange-looking head
SammyS
#2
Jul16-12, 09:41 PM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
Quote Quote by JJBladester View Post
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
#3
Jul17-12, 07:54 AM
PF Gold
JJBladester's Avatar
P: 287
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
#4
Jul17-12, 08:07 AM
Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,819
Finding resistor's voltage with current and voltage sources present

Quote Quote by JJBladester View Post
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
#5
Jul17-12, 08:56 AM
HW Helper
Thanks
P: 5,465
Quote Quote by JJBladester View Post
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.
JJBladester
#6
Jul17-12, 09:06 AM
PF Gold
JJBladester's Avatar
P: 287
Quote Quote by SammyS View Post
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:

[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 - VR2 - VR1 = 0
VR2 = E - VR1 = 4V - 1V = 3V


Register to reply

Related Discussions
How does voltage stay the same at all points in a circuit when a resistor is present? Introductory Physics Homework 4
Figuring out voltage across resistor and current in loop given initial voltage input Engineering, Comp Sci, & Technology Homework 2
Current and voltage sources Introductory Physics Homework 6
Finding current across a capacitor in a circuit with two AC voltage sources Introductory Physics Homework 13
Ideal voltage/current sources Electrical Engineering 3