Resistance of ammeter and voltmeter

In summary: Your problem here is that you assume that there is no current flow through the voltmeter (which there is).R2 is the equivalent resistance of the resistor and voltmeter; you're using just the resistor.Your problem here is that you assume that there is no current flow through the voltmeter (which there is).O, it's 1021 Ohms and not 1012 Ohms? Whew! I was going over my calculations trying to find where I went wrong.In summary, Homework Equations attempted to find the difference in current between circuits A and B by using the voltmeter. However, they were not able to find the answer correctly due to an assumption that there is no
  • #1
arkofnoah
124
1

Homework Statement


http://img638.imageshack.us/img638/2421/screenshot20100807at012.png [Broken]


Homework Equations





The Attempt at a Solution


For the ammeter resistance i chose to find the difference in the voltmeter reading in the two circuits. that difference is the p.d. across the ammeter, right?

but i don't know which ammeter reading to use now. in fact i think i have been on the wrong track all along.

perhaps someone can give some hint for me to get started?
 
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  • #2
arkofnoah said:
For the ammeter resistance i chose to find the difference in the voltmeter reading in the two circuits. that difference is the p.d. across the ammeter, right?

Yup :)


arkofnoah said:
but i don't know which ammeter reading to use now. in fact i think i have been on the wrong track all along.

perhaps someone can give some hint for me to get started?

Think of the current flow as water flowing. Draw the flow path in each figure.

How are resistance, current, and voltage related to one another?
 
  • #3
since circuit B has lower ammeter reading than circuit A, i can conclude that the difference in current is due to the current which flows through the voltmeter, right?

then from circuit B i know that the emf of the whole circuit is 10V (assume no internal resistance), and since the voltmeter is connected in parallel, the p.d. across it is 10V as well? with a bit of calculation i found that the resistance of the voltmeter is 1250 ohm. That is however not the answer (it's 1012 ohm by the way). What went wrong with my working here?
 
  • #4
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  • #5
arkofnoah said:
since circuit B has lower ammeter reading than circuit A, i can conclude that the difference in current is due to the current which flows through the voltmeter, right?

Yup!

arkofnoah said:
then from circuit B i know that the emf of the whole circuit is 10V (assume no internal resistance), and since the voltmeter is connected in parallel, the p.d. across it is 10V as well? with a bit of calculation i found that the resistance of the voltmeter is 1250 ohm. That is however not the answer (it's 1012 ohm by the way). What went wrong with my working here?
1012 Ohms is listed in your book? I'm getting 1023.5 Ohms for the resistance of the VM.

Does it get the amp meters resistance?
 
  • #6
arkofnoah said:
i have tried finding the resistance of ammeter in two ways. here is my working:

http://img36.imageshack.us/img36/9179/screenshot20100807at021.png [Broken]

i first did it using the potential divider principle. then i just used circuit B to find the resistance of the ammeter. i ended up with different answers, so which is the correct one?

Can you explain your work a little more for the top one? I'm not following what you did.
 
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  • #7
for the top one i just used the combined the two circuits into the potential divider arrangement akin to this:

[PLAIN]http://www.electronics2000.co.uk/images/calc/divider.gif [Broken]

treating the ammeter as another resistor. then i used the formula

[tex]V_{1} = \frac{R_{2}}{R_{2} + R_{1}} \times V_{2}[/tex]
after some simple derivations.

where V1 = 8.7V, V2 = 10V, R1 = resistance of ammeter and R2 = 200 ohm.

solving for this gives 29.9 ohm, which is wrong. the second one is correct, but why?
 
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  • #8
arkofnoah said:
for the top one i just used the combined the two circuits into the potential divider arrangement akin to this:

[PLAIN]http://www.electronics2000.co.uk/images/calc/divider.gif [Broken]

treating the ammeter as another resistor. then i used the formula

[tex]V_{1} = \frac{R_{2}}{R_{2} + R_{1}} \times V_{2}[/tex]
after some simple derivations.

where V1 = 8.7V, V2 = 10V, R1 = resistance of ammeter and R2 = 200 ohm.

solving for this gives 29.9 ohm, which is wrong. the second one is correct, but why?

Your problem here is that you assume that there is no current flow through the voltmeter (which there is).

R2 is the equivalent resistance of the resistor and voltmeter; you're using just the resistor.
 
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  • #9
6Stang7 said:
Your problem here is that you assume that there is no current flow through the voltmeter (which there is).

oh yea! it didn't occur to me. i just plugged in that formula from my textbook without realizing that the current must be the same for this formula to work.

so back to the voltmeter resistance, what am i doing wrong (i got 1250 ohm instead of 1021 ohm).
 
  • #10
O, it's 1021 Ohms and not 1012 Ohms? Whew! I was going over my calculations trying to find where I went wrong.

OK, can you show your work for calculating the voltmeter resistance?
 
  • #11
oh it was 1012 ohms. i just took 10V and divided it by the difference in ammeter readings (because the voltmeter "diverted" some current away in B while the ammeter reading in A is before this "diversion", so i suppose the difference is the current flowing through the voltmeter), but obviously it was wrong.
 
  • #12
arkofnoah said:
oh it was 1012 ohms. i just took 10V and divided it by the difference in ammeter readings, but obviously it was wrong.

O, hmmm; I get 1023.53 Ohms through a couple of different methods. Anyways...


Why does the amp read different values between the two circuits?
 
  • #13
6Stang7 said:
O, hmmm; I get 1023.53 Ohms through a couple of different methods. Anyways...Why does the amp read different values between the two circuits?

the voltmeter "diverted" some current away in B while the ammeter in A is placed before this "diversion", so i suppose the difference is the current flowing through the voltmeter
 
  • #14
That is true, but the total resistance of A is different then B, but the voltage across the equivalent circuit is the same (10V). Since V=IR, by changing the resistance, you change the current load.

Looking at Figure B, what can you tell me about the current through the resistor and through the voltmeter?
 
  • #15
ok i see. yea that's quite true the "resistors" are in different arrangement in both circuits so the current will naturally be different. failed to see that as well.

i manage to get the correct answer using circuit A. i think it's easier to use circuit A since you don't have that many resistors connected in parallel and the ammeter reading indicates the current for the entire circuit. so i just find out the equivalent resistance for the parallel circuit involving the voltmeter and the 200 ohm lamp using the reciprocal sum equation, get the ammeter and voltmeter reading, and with a bit of calculations i arrived at the answer.

but basically, problem solved. thanks for your help :smile:
 

1. What is the resistance of an ammeter and voltmeter?

The resistance of an ammeter and voltmeter is typically very low, with values ranging from 0.1 ohms to 100 ohms. This is to ensure that they do not affect the circuit being measured.

2. How do you calculate the resistance of an ammeter and voltmeter?

The resistance of an ammeter and voltmeter can be calculated by dividing the voltage or current range by the full-scale deflection. For example, if an ammeter has a full-scale deflection of 10 mA and a current range of 100 mA, the resistance would be 100 mA / 10 mA = 10 ohms.

3. Why is it important to consider the resistance of an ammeter and voltmeter?

The resistance of an ammeter and voltmeter can affect the accuracy of the measurement. If the resistance is too high, it can cause the circuit to be measured to behave differently than it would without the instrument. This can lead to inaccurate readings.

4. How does the resistance of an ammeter and voltmeter affect the circuit being measured?

The resistance of an ammeter and voltmeter can cause a small amount of voltage drop in the circuit being measured. This can be negligible for most circuits, but for more sensitive ones, it may need to be taken into account for accurate measurements.

5. Can the resistance of an ammeter and voltmeter be adjusted?

In most cases, the resistance of an ammeter and voltmeter cannot be adjusted. It is typically a fixed value determined by the instrument's design. However, some more advanced instruments may have the option to adjust the resistance for more accurate measurements.

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