Finding resistor's voltage with current and voltage sources presentby JJBladester Tags: current source, resistor, simulation, voltage source 

#1
Jul1612, 09:16 PM

PF Gold
P: 280

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? 



#2
Jul1612, 09:41 PM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,385

The difficulty appears to be that you are assuming the resistor, R_{2} in the original circuit has the same voltage across it as the resistor, R_{2}, in the last two circuits. 



#3
Jul1712, 07:54 AM

PF Gold
P: 280

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
Jul1712, 08:07 AM

Emeritus
Sci Advisor
HW Helper
PF Gold
P: 7,385

Finding resistor's voltage with current and voltage sources present 



#5
Jul1712, 08:56 AM

HW Helper
P: 4,706

Oops, my mistake. What you are using is not what I think of as current divider rule, but it's okay. 



#6
Jul1712, 09:06 AM

PF Gold
P: 280

Anyway, using your advice to find the voltage across R_{1} in the simplified circuit, I've found the voltage across R_{2} 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 


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