In this experiment, you will use a metre bridge to find the resistance of an unknown resistor X. You connects X and a standard 10.0 Ω resistor S in the gaps...
Question 1 Report
In this experiment, you will use a metre bridge to find the resistance of an unknown resistor X. You connects X and a standard 10.0 Ω resistor S in the gaps of the metre bridge as shown in Fig. 31.1. A cell and a galvanometer are also connected. You moves the sliding contact (jockey) along the bridge wire until the galvanometer reads zero. You records the balance length l1 from end A. You then swaps X and S and finds a new balance length.
You obtains the following results:
First reading: balance length l1 = 38.5 cm from end A.
After swapping X and S: balance length l2 = 61.0 cm from end A.
(a) Record the two balance lengths. [1]
(b) For the first reading, calculate X using X/S = l1/(100 − l1). Show your working. [2]
(c) For the second reading, calculate X using S/X = l2/(100 − l2). Show your working. [2]
(d) Calculate the average value of X from your two results. [1]
(e) State why you takes readings with X and S swapped. [1]
(f) State what the galvanometer reading of zero means. [1]
(g) Describe one precaution when using the jockey on the bridge wire. [1]
(h) State one source of error in this experiment. [1]
Answer Details
(a) The two balance lengths are: \( l_1 = 38.5 \; \text{cm} \) (first reading) and \( l_2 = 61.0 \; \text{cm} \) (after swapping X and S). [1]
(b) Using the balance condition for the first reading:
\[ \frac{X}{S} = \frac{l_1}{100 - l_1} = \frac{38.5}{100 - 38.5} = \frac{38.5}{61.5} = 0.626 \][1]
\[ X = 0.626 \times S = 0.626 \times 10.0 = 6.3 \; \Omega \][1]
(c) After swapping X and S, the balance condition becomes:
\[ \frac{S}{X} = \frac{l_2}{100 - l_2} = \frac{61.0}{100 - 61.0} = \frac{61.0}{39.0} = 1.564 \][1]
\[ X = \frac{S}{1.564} = \frac{10.0}{1.564} = 6.4 \; \Omega \][1]
(d) Average value of X:
\[ X_{\text{avg}} = \frac{6.3 + 6.4}{2} = 6.35 \; \Omega \][1]
(e) Swapping X and S and taking a second reading helps to eliminate systematic errors caused by non-uniform resistance along the bridge wire or by end corrections (extra resistance at the terminal connections). Averaging the two results gives a more reliable value. [1]
(f) A galvanometer reading of zero means there is no current flowing through the galvanometer. This occurs when the bridge is balanced: the potential difference across the two halves of the bridge are equal, so there is no driving force to push current through the galvanometer. [1]
(g) Do not press the jockey too hard onto the wire and do not drag it along the surface. Pressing too firmly can deform or scratch the wire, locally changing its cross-sectional area and resistance. Instead, tap the jockey briefly at each test position. [1]
(h) Sources of error include: contact resistance at the connections between the wire and the resistors, non-uniformity of the bridge wire (its resistance per unit length may vary along its length), or temperature changes in the wire caused by the current flowing through it. [1]
Why this matters: The metre bridge is a practical form of the Wheatstone bridge. At the balance point, the ratio of resistances in one arm equals the ratio in the other: \( \frac{X}{S} = \frac{l_1}{100 - l_1} \). Because the bridge wire has (ideally) uniform resistance per unit length, the resistance of each section is proportional to its length. The null method (zero galvanometer current) is powerful because at balance the measurement does not depend on the e.m.f. of the cell or the resistance of the galvanometer, eliminating two common sources of error.
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