Resistance of ammeter and voltmeter

[arkofnoah]

Homework Statement

http://img638.imageshack.us/img638/2421/screenshot20100807at012.png

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?

 

[6Stang7]

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 :)

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?

 

[arkofnoah]

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?

 

[arkofnoah]

i have tried finding the resistance of ammeter in two ways. here is my working:

http://img36.imageshack.us/img36/9179/screenshot20100807at021.png

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?

 

[6Stang7]

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!

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?

 

[6Stang7]

i have tried finding the resistance of ammeter in two ways. here is my working:

http://img36.imageshack.us/img36/9179/screenshot20100807at021.png

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.

 

[arkofnoah]

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

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

V_{1} = \frac{R_{2}}{R_{2} + R_{1}} \times V_{2}
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?

 

[6Stang7]

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

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

V_{1} = \frac{R_{2}}{R_{2} + R_{1}} \times V_{2}
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.

 

[arkofnoah]

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).

 

[6Stang7]

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?

 

[arkofnoah]

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.

 

[6Stang7]

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?

 

[arkofnoah]

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

 

[6Stang7]

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?

 

[arkofnoah]

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