# Calculating an unknown resistor

1. May 7, 2007

### Ry122

What do I have to do to calculate the value of the unknown resistor?
The galvonometer reads 0 AMPS.
I don't understand how the galvonometer changes according to where it's connected to the resistance wire. I thought amp meters read the same no matter where they're connected in the circuit.
Just for clarification, in the middle wire, the part of wire before the galvonometer is 3ohms and the part of wire after the galvonometer is 7 ohms.

2. May 7, 2007

### andrevdh

No current through the galvanometer implies that its leads are at the same potential, which means that Vx = V3 and V10 = V7.

3. May 7, 2007

### Four

Try using the loop rule along with the junction rule.

For the outer loop
$$-i_1 R - 10i_1 + 12V = 0$$
where i1 and i2 are the currents passing through unkown resistor and 10 Ohm resistor. Also R is the resistance of the unkown resistor.

Then the loop rule around 3 Ohm 7Ohm and the batery. Juction Rule at the connections of the Galvonometer. We have 5 unkowns and 4 equations.
2 loops 2 equations
2 juctions 2 equations
total 4

unkowns R and currents at each resistor: total 5

You can get your 5th equation by using the loop rule on the loop starting at batery moving through 3 Ohms, Galvonometer, 10 Ohms, back to battery.

I haven't tryed this out or looked more in it well. I hope this helps and the juctions are really 2 equations and not the same equation. Since resistance in the middle wire is 0 then its like just a ideal wire there.

Hope this helps
I'm off to work :)

4. May 7, 2007

### andrevdh

Unnecessary complicated:

$$\frac{V_7}{V_3} = \frac{V_x}{V_{10}}$$

which means that the resistances will be in the same ratio.

5. May 8, 2007

### Four

I never saw a way to solve a problem like this using a ratio of voltages. How does it work?

Thanks

6. May 8, 2007

7. May 8, 2007

### andrevdh

The left side of $$R_X$$ and the three ohm resistor are at the same potential, so are the right side of the ten and seven ohm resistors. Since the galvanometer conducts no current we know the its top and bottom leads are at the same potential. This means that the voltage drop over $$R_X$$ and the three ohm resistors are the same, ditto for that over the ten and seven ohm resistor. this means that the voltage ratios are as indicated in my previous post. Since

$$V = IR$$

we get that the current cancels out top and bottom on the left and right hand side of the equation (even though the current in the top and bottom branch differs) which leaves the resistance ratios

$$\frac{3}{7} = \frac{X}{10}$$

(I got the voltage ratios wrong in my previous post).

Last edited: May 8, 2007
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?