What is the Voltage read on a Real Voltmeter?

[tomrja]

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.

 

[berkeman]

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?

 

[tomrja]

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!

 

[berkeman]

Nice when that happens!

 

[rwisz]

I would like to clarify that a voltmeter is a device used to measure the potential difference or voltage between two points in an electrical circuit. It works by connecting in parallel with the component or circuit being measured, and the reading on the voltmeter represents the potential difference between those two points.

In this given scenario, the voltmeter V1 with an internal resistance of 4.80 kΩ is placed across R2, which is in series with the voltage source V0. To calculate the voltage reading on V1, we can use the voltage divider formula V1 = V0 * (R2 / (R1 + R2)). Substituting the given values, we get V1 = 32.4 V * (465 Ω / (205 Ω + 465 Ω)) = 24.3 V.

Similarly, for voltmeter V2 with an internal resistance of 2.75 MΩ, we can use the same formula V2 = V0 * (R2 / (R1 + R2)). Substituting the given values, we get V2 = 32.4 V * (465 Ω / (205 Ω + 465 Ω)) = 24.3 V.

In summary, the readings on both voltmeters V1 and V2 will be the same, which is 24.3 V. This is because the internal resistance of the voltmeters is much higher compared to the resistances of the circuit components, and therefore, it has a negligible effect on the overall circuit.