Potentiometer/Galvanometer problem

[thelibertine1]

Homework Statement

In the above (attached) potentiometer circuit the total range of the potential that can be measured by the 800ohm resistance AB is 1.6V. The resistance, R, of the meter G is 84 ohms. The slider is set at X to balance a potential E of 1.4000V.

a) What is the resistance included between X and B?
b) If the value of E is actually 1.4014V, what current will flow through the meter?
c) There are available other meters whose resistances,R, are different from that of the 84 ohm one, but which have the same senstivity for the same power supplied to them. What would be the optimum resistance which such a meter should have to make the above measurement(ie: what is the condition for maximum power transfer to the galvanometer?)

Homework Equations

V = IR

The Attempt at a Solution

The voltage between A and X is R1, X to B is R2
a)
I = the voltage across A to X / R1 + R2
I = 2/800 = 2.5E-3 A
voltage across R2 is then equal to voltage E ?
so R2 = V/I = 1.4 / 2.5E-3 = 560 ohms
Is this correct?

b) Not sure what they are asking to do. Not quite sure how the whole circuit works, is the current through G 0 in the first question? And is it not anymore? Why is the initial conditions important (the total range of potential) ?

c) Not sure where to start :S

Any help is really appreciated. Thanks.

 

[anathema]

Thank you for your post. I would like to offer some guidance on your questions.

a) Your calculation for the resistance between X and B is correct. The voltage across R2 is indeed equal to the voltage E, as the potentiometer is used to balance the potential difference between A and X with the potential difference between X and B. This means that the voltage across R1 will be equal to the voltage across R2, which is also equal to E.

b) The question is asking what the current through the meter G would be if the value of E is actually 1.4014V instead of 1.4000V. To answer this, we need to use Ohm's law, V = IR. Since the resistance of R2 is now 560 ohms (as calculated in part a), the current through the meter G would be I = V/R = 1.4014 / 560 = 2.50E-3 A. This means that the current through the meter G has increased by 0.50E-3 A compared to the initial conditions.

c) To find the optimum resistance for the meter G, we need to consider the condition for maximum power transfer. This condition is when the load resistance (in this case, the resistance of the meter G) is equal to the internal resistance of the source (in this case, the potentiometer). This means that the optimum resistance for the meter G would be 800 ohms, as this is the internal resistance of the potentiometer. This would ensure that the maximum amount of power is transferred to the meter G, allowing for the most accurate measurement.

I hope this helps. Let me know if you have any further questions.