Calculating Voltage Drop and Current Through a Voltmeter

In summary, the voltmeter has a voltage drop of 100v across the resistor, and the current through the voltmeter is 1mA.
  • #1
badd99
21
0

Homework Statement


Voltmeter has a internal resistance of 10M ohms and the resistor has a resistance of 100,000ohm. The current through the resistor is 1mA.
a)What is the voltage drop across the voltmeter?
b)What is the current through the voltmeter?


Homework Equations


V=IR is all you should need


The Attempt at a Solution


I am missing something really small (have taken the class already) but I'm stuck. I am stuck with how to apply the 1mA current. Do I need to use that to find the voltage of the source? Or do I find the overall resistance of the reistor and parallel voltmeter then use the 1mA somehow?

Please help me! Thank you!
 
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  • #2
You don't need to find the voltage of the source, or the combined resistance of the parallel resistor and voltmeter. Its simpler than that. You are given the resistance and current through the resistor, so what can you say about the voltage drop over the resistor? And what does this tell you about the voltage drop over the voltmeter?
 
  • #3
Well given 1ma over the resistor you can do V = (1E-3)(100,000) which is the voltage drop over the resistor...without the total voltage how can I find the voltage drop over the volt meter? current through voltmeter?
 
  • #4
Hint: Potential Difference across all parallel resistors is same.
 
  • #5
so if its V=(1E-3)(100,000) = 100 v across the resistor its 100 V across the voltmeter for part a?

For b you then take 100v=I(10E7) = 1E-6 A?
 
  • #6
Yes. Note that the voltage refers to potential difference and not always to source voltage
 
  • #7
badd99 said:
so if its V=(1E-3)(100,000) = 100 v across the resistor its 100 V across the voltmeter for part a?

For b you then take 100v=I(10E7) = 1E-6 A?

You got part (a) right. And I think you've got the right idea for part (b), but I think you went wrong in part (b). Check you have the right power which 10 is raised to. And write out the equations properly! It makes them easier for you and us to read when you write them out properly. (And your teacher will be able to read them easier, more importantly).
 
  • #8
I see my error, thank you guys very, very much for the help! Can't thank you enough!
 
  • #9
no worries :)
 

1. What is the purpose of a voltmeter?

A voltmeter is a device used to measure the voltage or potential difference between two points in an electrical circuit. It is commonly used to troubleshoot and diagnose electrical problems, as well as to ensure that the correct voltage is being supplied to a circuit.

2. How does a voltmeter work?

A voltmeter works by using a calibrated coil of wire, known as a galvanometer, which is connected in series with a known resistance. When a voltage is applied to the voltmeter, the current flows through the galvanometer, causing it to move. The amount of movement is then measured and converted to a voltage reading.

3. What is voltage drop over voltmeter?

Voltage drop over voltmeter refers to the decrease in voltage that occurs when a voltmeter is connected in series with a circuit. This is due to the internal resistance of the voltmeter, which causes a small amount of voltage to be dropped across the device.

4. How does voltage drop affect voltmeter readings?

Voltage drop can affect voltmeter readings by causing them to be slightly lower than the actual voltage in the circuit. This is because the voltmeter is measuring the voltage after a small amount has been dropped across its internal resistance. However, in most cases, this drop is negligible and does not significantly impact the accuracy of the reading.

5. How can voltage drop over voltmeter be minimized?

Voltage drop over voltmeter can be minimized by using a voltmeter with a lower internal resistance, using shorter and thicker leads when connecting the voltmeter to the circuit, and making sure the voltmeter is properly calibrated. It is also important to make sure the voltmeter is connected in parallel with the circuit, rather than in series, to avoid additional voltage drop.

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