# Resolution a problem with voltage divider

I'm facing a problem I don't exactly know how to resolve.

I have this circuit

The 2k resistance is a potentiometer. Basically we have a wheatstone bridge. Here's the picture

The question is : how do you adjust the potentiometer so that the amp meter shows a 0 current.

Well, the answer should be that - and only with reasoning on how the resistances on the two vertical branches (the lower/upper part of the potentiometer) act as a tension divider, or at least that's how they did it in my solution book - the upper part (say X) and lower part (say Y) of the potentiometer should make it so that X/Y = 8,2/3.

Seriously I don't get it. I have tried methods using superposition principle (considering for example X in serial with Req(Y+3k), and 8,2 in serial with Req(Y+3k)), but I end up with X = 8,2k which is non sense since the potentiometer is 2k.

Can someone explain? I know since its an amp meter it's just like if it was not there. I don't get how you can use the tension divider law if the the middle of the two vertical branches are connected with a wire (since I consider the amp meter as a wire, "ideal amp meter").

• You know X/Y = 8.2/3 (assuming that's correct). You also know that X+Y=2. With that, you can solve X and Y. – RJR Oct 24 '13 at 5:41
• Actually, their conclusion IS that the relation between X/Y = 8.2/3. I shouldn't know it, at least until I checked the solutions. And I will be dread honest with you, I still don't get, with that, how Ic should be 0. When I check it with a glance, only X = 8,2 and Y = 3 should make Ic = 0. Guess I'm mistaken... I'm clueless here honestly. – Yannick Oct 24 '13 at 5:47
• Measure the 8.2k to 3k junction voltage, and the 2k wiper voltage. When zero voltage difference, zero current flow. Highest sensitivity is when all 4 legs are equal to each other, and your Ammeter is a very high impedance Voltmeter, or even the voltage between a pair of voltage followers. – Optionparty Oct 24 '13 at 6:30

I built this in CircuitLab.

Consider the sliding contact of the potentiometer to divide the 2kΩ resistance into two parts, the part "above" the sliding contact and the part "below" the sliding contact.

Set the position of the sliding contact so that the ratio of the resistance above the contact to the resistance below the contact is 8.3:3. This gives resistances of $2kΩ \times \frac{8.3}{3.0+8.3} = 1,464Ω$ and $2kΩ \times \frac{3.0}{3.0+8.3} = 536Ω$.

When simulated, the current through R2 is 13 mA and the current through R6 is 23 uA. That is, the current through R6 is a thousand times less than the current through R2, i.e. effectively zero.

Note that $R2/R3 = R4/R5$ so that the voltages at "NODE 2" and "NODE 3" are identical, so that there is no potential difference between "NODE 2" and "NODE 3" and the current must necessarily be zero.

If you need an analytical solution, just replace the ampermeter with a voltmeter. Then solve the schematic, for 0 voltage across the ends of the voltmeter. In these conditions, if you replace the voltmeter with ampermeter, the current will remains 0.

There are only two possible conditions for current not flowing in the ammeter: -

• The 45V battery is removed (stupid of course)
• The voltage on the wiper being the same as the 8k2 and 3k resistors

If, the 8k2 and 3k resistors produce X volts due to the ratio $\dfrac{3k}{8.2k+3k}$, then the pot has to be set so that the ratio of its lower part (the resistance below the wiper) to 2k ohms is the same.