Potential difference across a resistor as measured by a voltmeter.

In summary, the voltmeter shown in the figure can be modeled as an ideal voltmeter with infinite internal resistance in parallel with an 11.0-MΩ resistor. The reading on the voltmeter can be calculated using the equation V = EMF*R/3R, where EMF is the electromotive force and R is the resistance. This equation holds true for all three given values of R (1.1 MΩ, 10.8 MΩ, and 104 MΩ) as long as the 11.0-MΩ internal resistance is taken into account.
  • #1
kballerina
1
0

Homework Statement



The voltmeter shown in the figure below can be modeled as an ideal voltmeter (a voltmeter that has an infinite internal resistance) in parallel with a 11.0-MΩ resistor. Calculate the reading on the voltmeter for the following.

25-p-091.gif



R = 1.1 MΩ


R = 10.8 MΩ


R = 104 MΩ

Homework Equations



V=IR
Rtotal=R+2R (since they are in series)

The Attempt at a Solution



My guess is that I am having two issues with this problem, the first being the problem statement itself, it gives me a value for a R and then tells me to solve the problem for the given values of R. I got the first two parts of the problem ( R = 0.9 kΩ and R = 9.7 kΩ both having the potential difference of 3.333 volts) To do those I divided the EMF by Rtotal (since the resistors are in series) to get the current then multiplied by the resistance of R. This worked fine for the first two but on the third one and beyond my problem arose and my answer was wrong. I'm not entirely sure why and the explanation in my book is quite limited so I'm having trouble finding my error on these last 3 parts of the problem. For all of my equations the answer was same potential difference (3.333 V)
 
Last edited:
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  • #2
Hi kballerina, welcome to PF.

Reading in the voltmeter is given by

V = EMF*R/3R.

And it is true for all the three cases.
 
  • #3
rl.bhat said:
Hi kballerina, welcome to PF.

Reading in the voltmeter is given by

V = EMF*R/3R.

And it is true for all the three cases.

You're forgetting the 11 M[tex]\Omega[/tex] internal resistance of the voltmeter.
It's parallel with resistance R.
 

Related to Potential difference across a resistor as measured by a voltmeter.

1. What is potential difference?

Potential difference, also known as voltage, is the difference in electric potential energy per unit charge between two points in an electric circuit. It is measured in volts (V).

2. How is potential difference across a resistor measured?

Potential difference across a resistor can be measured using a voltmeter, which is connected in parallel to the resistor. The voltmeter measures the potential difference between the two points in the circuit, also known as the voltage drop.

3. What is the role of a resistor in potential difference?

A resistor is a component that resists the flow of electric current in a circuit. As a result, it causes a potential difference across itself, which can be measured by a voltmeter. The higher the resistance of a resistor, the greater the potential difference across it.

4. Can potential difference across a resistor be negative?

Yes, potential difference can be negative across a resistor if the voltmeter is connected in reverse polarity. This means that the positive lead of the voltmeter is connected to the negative side of the resistor and vice versa. In this case, the voltmeter will display a negative value, indicating a reverse potential difference.

5. How does the potential difference across a resistor affect the flow of electric current?

The potential difference across a resistor is directly proportional to the flow of electric current. This means that as the potential difference increases, the current also increases. However, this relationship is affected by the resistance of the resistor, as higher resistance will limit the flow of current even with a higher potential difference.

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