Question
The question are attached. The questions are two which is down in this page. Please write the answer in a paper and attach it and you will recieve the full points.
This experiment does not require high voltage source. It is recommended that approximately 5V. Cheek that the apparatus is set up as indicated in Figure 1.Set Rx (the Heathkit resistance substitution box) to 4.7 kohm. We begin with RG 5 kohm; and as a more precise balance is obtained, we will decrease it gradually to zero. An attempt should be made to balance the bridge as close to the 50-cm mark along the slide wire as possible. You will now determine the value of the unknown resistance Rx. Initially, the point contact C should be made at the 50 cm mark so that R2 = R3 (L2 = L3). R1 should be adjusted as accurately as possible to obtain a balance point. Tap the wire for a fraction of a second and note the amount of deflection of the galvanometer needle. Change the value of R1 and tap once again ) to see if the galvanometer deflects the same way or the opposite way. Continue the procedure until you succeed in finding two values of the resistance R1 that give oppositely directed galvanometer deflections. The balance point must lie between these values. Remember, RG, should gradually be decreased as liner balancing is achieved. When the adjustment of R| is completed, vary the slice point C to obtain a still more accurate balance. For your final balance, make RG = 0 by placing a lead across the terminals of the resistance substitution box. Record the value of R1, L2, and L3. Calculate a value for the unknown resistance. Be sure to reset RG = 5 kohm before measuring the next resistance. Measure the resistance of the Heathkit resistance substitution box following the procedure described above. Determine the percent error between the actual resistance value and your experimental one. Reset Rx (the Heathkit resistance substitution box) to 470ohm and repeat step 1. Measure the resistance between the terminals 2 and 3, or 3 and 4, or 4 and 5, on the unknown resistance spool as in Step I. The type and length of wire of which the spool vs made is indicated. The actual resistance value of the spool can be determined by obtaining values for p, the cross-sectional area A from you instructor, and by using equation (I). Compare this to the value determined experimentally. Report any broken equipment to your instructor. Clean up your work area when finished. 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 measuring L2 and L3 depends on this width.) In your own words, explain why the Wheatstone Bridge experiment is both a null and comparison method.
Explanation / Answer
1. because the probe by which you are dividing the wire to two different parts....L1 and L2 has finite leangth and is short. So it is easy to measure the leangths near %0 cm mark accurately because the couldn't reach both ends more precisely and we need some accurate readings. It's because of the easy experiment proceeding (according to experimental set=up)
2. There are two principal ways of operating a bridge such as this. One is by operating it as a
null detector, where the bridge measures resistance indirectly by comparison with a
similar standard resistance. On the other hand, it can be used as a device that reads a
resistance difference directly, as a proportional voltage output.
When R1/R4 = R2/R3, the resistance bridge is said to be at a null, irrespective of the
mode of excitation (current or voltage, AC or DC), the magnitude of excitation, the mode
of readout (current or voltage), or the impedance of the detector. Therefore, if the ratio of
R2/R3 is fixed at K, a null is achieved when R1 = K
