Why Does V_R3 Measure 100 V in a Mixed Resistor AC Circuit?

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SUMMARY

The discussion clarifies the voltage measurements in a mixed resistor AC circuit, specifically addressing the confusion regarding V_R3. The calculations confirm that V_R1 equals 20 V and V_R2 equals 100 V, leading to the conclusion that V_R3 is indeed 100 V, not 0 V. This is due to the principle that voltage across resistors in parallel remains the same, which is a fundamental difference from DC circuits where voltages add up. The correct understanding of voltage drops in parallel circuits is essential for accurate analysis.

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Brad1983
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I am confused when it comes to this AC circuit problem. It is part e. The back of the book shows that V_R3 = 100 V. But as you see from my calculations, I get V_R1 = 20 V and
V_R2 = 100 V. So shouldn't V_R3 = 0 V because the original voltage source for this AC circuit is 120 V, yet the book says V_R3 = 100 V.

I am used to DC where you add up the voltages on each resistor and see if it equals the total voltage source. Hope you guys can help because we just started this stuff this week.

Q: Draw the diagram of a three resistor series parallel circuit having a 10-kΩ resistor (labeled R_1) in series with the source, and two 100-kΩ resistor (labeled R_2 and R_3) in parallel with each other and in series with R_1. Assume a 120-V ac source. Calculate the following parameters and appropriate label them on the diagram:
a.) V_R1
b.) I_R1
c.) P_R1
d.) I_R2
e.) V_R3
f.) ɸ between V applied and I total
__________________________________________________ _________________________
My Calculation:
Picture of Circuit:
http://img87.imageshack.us/f/circuith.png/

R_equ1 = 1/((1/R_2)+(1/R_3)) = 1/((1/100 kΩ)+(1/100 kΩ)) = 50 kΩ
R_T = R_1 + R_equ1 = 10 kΩ + 50 kΩ = 60 kΩ
V_T = I_T * R_T
I_T = V_T / R_T = 120 / 60 kΩ = 2 mA

a.)
V_R1 = I_T * R_1= 2 mA * 10 kΩ = 20 V
b.)
V_R1 = I_R1 * R_1
I_R1 = V_R1/ R_1 = 20 V / 10 kΩ = 2 mA
c.)
P = V_R1² / R_1 = (20 V)² / 10 kΩ = 40 mW
d.)
V_T = V_R1+ V_R2
V_R2= V_T - V_R1= 120 V – 20 V = 100 V
V_R2= I_R2* R_2
I_R2 = V_R2/ R_2 = 100 V / 100 kΩ = 1 mA
e.)
V_T = V_R1 + V_R2 + V_R3
V_R3= V_T - V_R1- V_R2= 120 V – 20 V – 100 V
= 0V
 
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Welcome to PF!

Hi Brad1983! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)
Brad1983 said:
… I get V_R1 = 20 V and
V_R2 = 100 V. So shouldn't V_R3 = 0 V because the original voltage source for this AC circuit is 120 V, yet the book says V_R3 = 100 V.

voltage only adds in series, voltage in parallel is the same

in this case the voltage drop (ie, the difference in potential) across R2 is the same as across R3

in both cases, it's the difference in potential between those two junctions in the middle of the diagram :smile:

(always think of voltage drops, or voltage differences, between two points, rather than just voltage :wink:)
 


tiny-tim said:
Hi Brad1983! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)


voltage only adds in series, voltage in parallel is the same

in this case the voltage drop (ie, the difference in potential) across R2 is the same as across R3

in both cases, it's the difference in potential between those two junctions in the middle of the diagram :smile:

(always think of voltage drops, or voltage differences, between two points, rather than just voltage :wink:)


I see, so it would stay the same regardless...I forgot about that.

Thanks Tim.
 

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