What is the Voltage read on a Real Voltmeter?

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  • #1
tomrja
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Homework Statement


The output of the voltage-divider network shown in the diagram below is to be measured with two different voltmeters, V1 and V2. Consider the situation when V0 = 32.4 V, R1 = 205 Ω, and R2 = 465 Ω.

a) When the voltmeter V1, whose internal resistance is 4.80 k Ω is placed across R2, what would be the reading on the voltmeter V1?

b) When the voltmeter V2, whose internal resistance is 2.75 M Ω, is placed across R2, what would be the reading on the voltmeter V2?

Homework Equations



1/Req=(1/R1)+(1/R2) where Req= equivalent resistance

Req=R1+R2

V=IR


The Attempt at a Solution


I really do not know where to start with this problem. I originally found the equivalent resistance using the resistors in parallel equation and used that resistance to calculate the voltage that was running through that resistor. Using Kirchoff's Rule that the equation for the circuit was V-R1*I-Req*I=0 where Req in this case is the equivalent resistance of the voltmeter and R2. I guess am having trouble understanding what the voltmeter actually does to the circuit.
 

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Answers and Replies

  • #2
berkeman
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Homework Statement


The output of the voltage-divider network shown in the diagram below is to be measured with two different voltmeters, V1 and V2. Consider the situation when V0 = 32.4 V, R1 = 205 Ω, and R2 = 465 Ω.

a) When the voltmeter V1, whose internal resistance is 4.80 k Ω is placed across R2, what would be the reading on the voltmeter V1?

b) When the voltmeter V2, whose internal resistance is 2.75 M Ω, is placed across R2, what would be the reading on the voltmeter V2?

Homework Equations



1/Req=(1/R1)+(1/R2) where Req= equivalent resistance

Req=R1+R2

V=IR


The Attempt at a Solution


I really do not know where to start with this problem. I originally found the equivalent resistance using the resistors in parallel equation and used that resistance to calculate the voltage that was running through that resistor. Using Kirchoff's Rule that the equation for the circuit was V-R1*I-Req*I=0 where Req in this case is the equivalent resistance of the voltmeter and R2. I guess am having trouble understanding what the voltmeter actually does to the circuit.

The voltmeter presents an equivalent parallel output resistance to the circuit. So the lower the output resistance of the voltmeter (bad), the more it pulls down the voltage being measured in the middle of the voltage divider.

What voltage did you get for the two different cases of the parallel voltmeter output impedance? That's pretty much your answer.

Model the voltmeter as an infinite input impedanc amplifer (that measures the voltage), in parallel with its output resistance. The output resistance "loads" the circuit being measured. So the higher the output resistance of the meter, the more accurate the reading will be (the closer to the unloaded circuit).

Make sense?
 
  • #3
tomrja
18
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The voltmeter presents an equivalent parallel output resistance to the circuit. So the lower the output resistance of the voltmeter (bad), the more it pulls down the voltage being measured in the middle of the voltage divider.

What voltage did you get for the two different cases of the parallel voltmeter output impedance? That's pretty much your answer.

Model the voltmeter as an infinite input impedanc amplifer (that measures the voltage), in parallel with its output resistance. The output resistance "loads" the circuit being measured. So the higher the output resistance of the meter, the more accurate the reading will be (the closer to the unloaded circuit).

Make sense?

The part about how the higher resistance of the voltmeter makes sense because the inverse property of adding resistors in parallel.

Here are the numbers that I worked out. Maybe you can tell me where I went wrong.

I first found the equivalent resistance of the voltmeter and both resistors and found that to be 628.9316ohms. Then I calculated the current flowing through the circuit using I=V/R=(32.4V)/(628.9316ohms)=0.0515159A. Then, using that current I used V=IR once again but this time I used the resistance between the voltmeter and R2 and got V=(0.0515159A)(423.9316ohm)=21.84V. Wait... haha I got it! I guess it just took you explaining a little for me to figure it out. Thanks!!
 
  • #4
berkeman
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63,238
14,189
Nice when that happens!
 

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