# A Multirange Coil Ammeter Galvonmeter (?) resistance question

[SOLVED] A Multirange Coil Ammeter Galvonmeter (???) resistance question

## Homework Statement

A Multirange Ammeter. The resistance of the moving coil of the galvanometer G in the figure (See Below) is 48 Ohms, and the galvanometer deflects full scale with a current of 0.0200 A. When the meter is connected to the circuit being measured, one connection is made to the post markedand the other to the post marked with the desired current range.

Part A

Find the magnitude of the resistance R1 required to convert the galvanometer to a multirange ammeter deflecting full scale with currents of 10.0 A, 1.00 A, and 0.100 A.

Part B

Find the magnitude of the resistance R2 required to convert the galvanometer to a multirange ammeter deflecting full scale with currents of 10.0 A, 1.00 A, and 0.100 A.

Part C

Find the magnitude of the resistance R3 required to convert the galvanometer to a multirange ammeter deflecting full scale with currents of 10.0 A, 1.00 A, and 0.100 A.

## Homework Equations

(I'm not sure which are actually relevant, so I will put a few down)

P=IV

V=IR

Kirchoff Rules:

Current Going into Junction equals total current Out of Junction

Total Voltage over a loop = 0

Resistors in Series:

$$R_{equv} = R_1 + R_2 + R_3$$

## The Attempt at a Solution

The question has confused me as to what I should do!!! Are we supposed to assume that the other resistors aren't there when find the Resistance?

Any Ideas will be greatly appreciated,

TFM ( )

Related Introductory Physics Homework Help News on Phys.org
alphysicist
Homework Helper
Hi TFM,

I did not see a diagram, so I'm not sure how these are all hooked up; perhaps I'm thinking of the problem setup completely wrong! However, I think the general idea is that you have resistors hooked up along with your galvanometer and that entire setup (galvanometer+resistor) is your ammeter.

Then, if you want the ammeter to deflect full scale at 10 A, you choose the resistors so that when 10 A enters the ammeter, only 0.02A enters the galvanometer coil.

If this doesn't answer your question, perhaps you could post a picture of the circuit or give more description?

I meant to put the image on , Sorry, Anyway, here it is, from the MP Question

TFM

#### Attachments

• 11.5 KB Views: 657
Any ideas now that the there is the diagram?

TFM

alphysicist
Homework Helper
My last post had the idea to get the resistances of the three resistors. Did you try it? The point was that, for example when being used as a 10 A ammeter, you know the current through the galvanometer and r2 and r3 are all the same (because they are in series); you also know the numerical values of this current and therefore also the current through r1.

Writing a Kirchoff loop equation for all three situation should give you your answer.

I still seem to be slightly confused (Sorry)

When we need to work out the resistance f R1, what happens to the resistanmce in R2 and R3? Also, we are not given a voltage?

TFM

alphysicist
Homework Helper
I believe the way the question is written is rather confusing. It seems to be suggesting that you'll find three resistance values for all of the resistors, but I do not believe that is correct; also, you can't find R1 by itself. Depending on which of the terminals you are using (10A, 1A, or 0.1A) you essentially have three different circuits.

For the 10 A circuit, just ignore the 1A and 0.1A leads. This means you have the galvanometer, R2, and R3 in series, and they are in parallel with R1. For the case of maximum deflection, you know the currents through each branch. If you write a Kirchoff's loop equation (on just the loop), you don't need the voltage. (Although you can find the maximum voltage for the 10A case right away with Ohm's law.) What loop equation do you get for the 10A case?

Using only the 1A lead gives a different circuit, and a different loop equation; and the same for the 0.1A lead. With three equations and three unknowns, you can solve for all three at once. What do you get?

Would these diagrams look right to find the first Resistance. If so, would the first equation be:

$$10R_1 + I(R_2 + R_3) + 48I_{Galvometer}$$

TFM

#### Attachments

• 191.5 KB Views: 530
• 213.5 KB Views: 486
• 179.3 KB Views: 452
alphysicist
Homework Helper
Would these diagrams look right to find the first Resistance. If so, would the first equation be:

$$10R_1 + I(R_2 + R_3) + 48I_{Galvometer}$$

TFM
I can't see the attachments yet, but that's the right kind of idea for the first; but there are several problems.

The current is going to enter the ammeter through the + terminal and leave through the one marked 10A. The current in each branch will therefore go to the right through r1 and the galvanometer in the diagram. So as you follow the loop around, when you go against the current the voltage difference will be negative. So if you write your loop equation down clockwise, the IR term for R1 will be negative. (If you go counterclockwise, the other three terms will be negative.)

Also, there are 10A coming into the + terminal, but that splits up in the two branches. We want the galvanometer to be at full deflection when 10A comes in, and we know what current in the galvanometer gives full deflection (it's given in the problem statement). So that's the current in the galvanometer, and it's also the current in R2 and R3 when its set up to be a 10A ammeter. What is that current?

So if 10A is coming in, and you know how much is going into the galvanometer branch, how much is going through the branch with R1?

Once you make those changes and set your expression to zero, you have an equation with three unknowns. What do you get?

Also, there are 10A coming into the + terminal, but that splits up in the two branches. We want the galvanometer to be at full deflection when 10A comes in, and we know what current in the galvanometer gives full deflection (it's given in the problem statement). So that's the current in the galvanometer, and it's also the current in R2 and R3 when its set up to be a 10A ammeter. What is that current?
Would the current be 0.020 Amps?

So if 10A is coming in, and you know how much is going into the galvanometer branch, how much is going through the branch with R1?
Using Kirchoffs Law, current in = Current out,

would the current be 10 - 0.02 = 9.98 Amps

Would the changes make the equation:

$$-9.98R_1 + 0.02(R_2 + R_3) + 480.02$$

Edit: Equation should have been:

$$-9.98R_1 + 0.02(R_2 + R_3) + 48*(0.02)$$

Sorry

TFM

(I have reattached the diagrams in a more visible form - I didn't realise Bitmap images didn't show well!)

#### Attachments

• 7 KB Views: 555
• 7.4 KB Views: 518
• 7.6 KB Views: 474
alphysicist
Homework Helper
TFM,

Your final equation for the 10A case (once you set it to zero) looks right to me.

Did you create those diagrams, or did they give them to you?

The way I was thinking of designing it was, for the 10A case, just ignore the 1A and 0.1A leads. They won't be connected to anything, so they are not affecting the circuit. If you are drawing a picture, just erase them!

Do the same for the 1A case; you're connecting this device to a circuit using only the + lead and the 1A lead; then the 10A and 0.1A lead are not doing anything. In that case, the galvanometer is in series with R3; also R1 and R2 are in series with each other; and (galvanometer+R3) is in parallel with (R1+R2). What loop equation do you get for the 1A case?

Yeah I drew them myself, but what you said about ignoring the other leads makes more sense.

So:

$$-9.98R_1 + 0.02(R_2 + R_3) + 48*(0.02) = 0$$

So for circuit 2:

$$0.02(48 + R_3) - (1-0.02)(R1 + R2) = 0$$

And for Circuit 3:

$$0.02(48) - (0.1-0.02)(R_1+R_2+R_3)$$

Do these equations look right?

TFM

alphysicist
Homework Helper
The loop equations look right to me; they are all equal to zero and now it's a matter of algebra.

I have rearranged the equations to give:

$$0.02R_2 + 0.02R_3 + 0.96 = 9.98R_1$$ - (1)

$$R_3 = 49R_1+49R_2 - 48$$ - (2)

$$0.96-0.92R_1 - 0.92R_3 = R_2$$ - (3)

Putting (2) into (3)

$$R_2 = -0.9375 - 0.985R_1$$ - (4)

PUtting (4) into (1):

$$0.9412 + 0.02R_3 = 9.96R_1$$ - (R1.5)

putting (3) into (2):

$$R_3 = 499R_1 - 470.6$$ - (5)

Putting (5) into (1.5)

$$R_1 = 423.54$$

This seem slightly high, though?

Edit: and according to MP, is wrong?

TFM

alphysicist
Homework Helper
I have rearranged the equations to give:

$$0.02R_2 + 0.02R_3 + 0.96 = 9.98R_1$$ - (1)

$$R_3 = 49R_1+49R_2 - 48$$ - (2)

$$0.96-0.92R_1 - 0.92R_3 = R_2$$ - (3)
I think this Eq. 3 is incorrect.

Does this look better?

$$0.077-0.0064R_1 - 0.0064R_3$$ - (3)

TFM

alphysicist
Homework Helper
Does this look better?

$$0.077-0.0064R_1 - 0.0064R_3$$ - (3)

TFM
I don't think so, because in your last post, you were trying to solve Eq. 3 for R2. What is this expression equal to? I don't think it is equal to R2. Remember that the original equation for circuit 3 is:

$$0.02(48) - (0.1-0.02)(R_1+R_2+R_3)=0$$

I would combine all of the numerical factors first and then solve for R2.

Lets See:

$$0.02(48) - (0.1-0.02)(R_1+R_2+R_3) = 0$$

$$(0.96) - (0.08)(R_1+R_2+R_3) = 0$$

$$(0.96) - (0.08R_1 + 0.08R_2 + 0.08R_3) = 0$$

$$(0.96) - 0.08R_1 - 0.08R_2 - 0.08R_3 = 0$$

$$(0.96) - 0.08R_1 - 0.08R_3 = 0.08R_2$$

$$12 - R_1 - R_3 = R_2$$

Is this better?

TFM

alphysicist
Homework Helper
Yes, that's looks right.

Okay, so:

Put (3) into (2):

$$R_3 = -2.04R_1 - 0.075$$ (4)

put (4) into (3):

$$12.075 + 1.04R_1 = R_2$$ (5)

Put (5) and (4) into (1)

$$0.02(-2.04R_1 - 0.075 ) + 0.02(12.075 + 1.04R_1 ) + 0.96 = 9.98R_1$$

Gives me

R_1 = 0.12 Ohms

Does this look right?

TFM

alphysicist
Homework Helper
That looks right to me; probably the best thing to do is to go ahead and solve for r2 and r3 from Eq. 4 and Eq. 5, and then plug them into the original three equations to see if they are solutions (to check for any algebraic errors).

I htink I may have an error somewhere because I get:

R2 = 12.2

R3 = -0.3

You can't get a negative resistance, surely?

TFM

alphysicist
Homework Helper
I think the answer to r1 is right, so let's look at the others.

Okay, so:

Put (3) into (2):

$$R_3 = -2.04R_1 - 0.075$$ (4)
I don't think this is right. Eq. 3 is:

$$12 - R_1 - R_3 = R_2$$

and Eq. 2 is:

$$R_3 = 49R_1+49R_2 - 48$$

If you plug Eq. 3 into Eq.2, you get a nice result: r1 and r2 immediately cancel from the equation and so you can find r3 without anything else. If you're not getting that result, try rewriting them like this:

Eq. 3:

$$(R_1 + R_2) = 12-R_3$$

and Eq. 2:

$$R_3= 49 (R_1+R_2) - 48$$

so you just plug (12-r3) into the appropriate place in this last Eq. 2 and solve for r3. The answer should be positive. What do you get?

Once you have r3, you can plug it back into the new eq. 3 above and solve for r2.

Thanks, I have now got R2 = 1.08 and R3 - 10.8. And inserting them into MP shows they are correct.

Thanks very much, alphysicist TFM