1) why is it beneficial to balance the wheatstone bridge as closely as possible

Question

1) why is it beneficial to balance the wheatstone bridge as closely as possible to the 50-cm mark along the slide wire? (hint : the probe has a finite width and the accuracy of measureing L2 and L3 depends on this width.)

2) to determine an unknown resistance , which is more accurate: ohm's law or wheatstone bridge method? explain your answer.

3) does it matter if the current supplied by the power supply is not constant?

Procedure This experiment does not require high voltage source. It is recommended that you use approximately 5V. Check that the apparatus is set up as indicated in Figure I. Set Rx (the Heathkit resistance substitution box) to 4.7 k ohm. We begin with RO= 5ohm; and as a more precise balance is obtained, we will decrease it gradually to zero. An attempt should he made to balance the bridge as close to the 50-cm mark along the slide wire as possible. 1) why is it beneficial to balance the wheatstone bridge as closely as possible to the 50-cm mark along the slide wire? (hint : the probe has a finite width and the accuracy of measureing L2 and L3 depends on this width.) 2) to determine an unknown resistance , which is more accurate: ohm's law or wheatstone bridge method? explain your answer. 3) does it matter if the current supplied by the power supply is not constant?

Explanation / Answer

1)

It is beneficial to balance the wheatstone bridge as closely as possible because, at balance, no current flows through the galvanometer, so the potential difference across it is zero. This implies that the current flowing through the known resistor P must be equal to that going through Q (P and Q are fixed resistors). In the same way, the current throughout the wire is constant (although different to that passing through the resistors). As the potential diff. across the galvanometer is 0, the voltages are constant and you can use Ohm's law to prove that the following equation holds.

As V = IR Volate = V is constant and The I1 current is that flowing through P and Q, the I2 current is that going through l1 and l2

I1 P = I2 L1 and I1Q = I2 L2

On diving we get,

P :Q = L1 :L2

The metre bridge obeys this relationship. By moving the jockey key, you change the ratio of l1:l2, so when the galvanometer reads 0 the ratio of the lengths is the same as the ratio of the fixed resistors. Resistance in the wire is directly proportional to it's length because it is uniform, that is, it has a constant cross-sectional area and so the resistivity equation

R= (rho)L/A (Rho and A are constants)

Also, D = diameter = 2 x sqrt ( rho l L1 / pi R L2)

2)

using Ohm

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