Finding resistor's voltage with current and voltage sources present

In summary, the voltage across R2 can be found by first converting the 4 V source and resistor into a current source, I_2, and combining it with the existing current source, I_1. Then, using the current divider rule, the voltage across R2 (I_R2) can be found. However, this voltage may differ from the voltage across R2 in the original circuit, which can be found by using Kirchhoff's Voltage Law and the current divider rule on the simplified circuit.
  • #1
JJBladester
Gold Member
286
2
What is the voltage across R2? You should convert the 4 V source and resistor into a current source as your first step.

current_and_voltage_source.png


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:

current_and_voltage_source_2.png


We can combine the parallel current sources I_1 and I_2:

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

current_and_voltage_source_3.png


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?
 
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  • #2
JJBladester said:
What is the voltage across R2? You should convert the 4 V source and resistor into a current source as your first step.

current_and_voltage_source.png


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:

current_and_voltage_source_2.png


We can combine the parallel current sources I_1 and I_2:

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

current_and_voltage_source_3.png


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.
 
  • #3
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.
 
  • #4
JJBladester said:
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.
 
  • #5
JJBladester said:
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
[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 [Broken]
Oops, my mistake. What you are using is not what I think of as current divider rule, but it's okay.
 
Last edited by a moderator:
  • #6
SammyS said:
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:

kvl.jpg


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
 

1. What is the formula for finding a resistor's voltage with current and voltage sources present?

The formula for finding a resistor's voltage when both current and voltage sources are present is V = IR, where V is the voltage across the resistor, I is the current flowing through the resistor, and R is the resistance of the resistor.

2. How do I determine the polarity of the voltage across the resistor?

The polarity of the voltage across the resistor can be determined by using the passive sign convention. If the current is flowing in the same direction as the voltage drop across the resistor, the polarity will be positive. If the current is flowing in the opposite direction, the polarity will be negative.

3. Can I use Ohm's Law to find the voltage across a resistor with both current and voltage sources present?

Yes, Ohm's Law can be used to find the voltage across a resistor even when both current and voltage sources are present. This is because Ohm's Law states that the voltage across a resistor is directly proportional to the current flowing through it.

4. What is the difference between a current source and a voltage source?

A current source is a type of electronic circuit that supplies a constant amount of current, regardless of the voltage across it. A voltage source, on the other hand, supplies a constant amount of voltage, regardless of the current flowing through it.

5. How does the presence of both current and voltage sources affect the voltage across a resistor?

The presence of both current and voltage sources can affect the voltage across a resistor in different ways. If the current and voltage sources are in series, the voltage across the resistor will be the sum of the voltage drops across each source. If they are in parallel, the voltage across the resistor will be equal to the voltage of the voltage source. It is important to analyze the circuit carefully to determine the exact voltage across the resistor.

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