Explain with a neat circuit diagram, how you will determine the unknown resistance by using a meter-bridge.

Question

Accepted Answer

A metre bridge (Section 9.3.1) is set up with its one-metre resistance wire AB stretched along a metre scale, its ends soldered beneath two L-shaped metal strips, and a third strip between them splitting the top into a left gap and a right gap. The unknown resistance X is connected across the left gap and a resistance box across the right gap; a galvanometer's one terminal is wired permanently to the central strip, its other terminal ending in a jockey that is tapped along the exposed wire. A suitable resistance R is chosen on the box, and the jockey is tapped at different points until a null point D is found where the galvanometer shows no deflection. Writing $l_x$ for the wire-length between end A and D, and $l_R=100-l_x$ for the remaining length to end B, the balance condition (from the underlying Wheatstone-bridge balance, with the two wire segments' resistances $R_{AD}=\rho l_x/A$ and $R_{DB}=\rho l_R/A$ standing in for two of the four bridge arms) is $X/R = R_{AD}/R_{DB} = l_x/l_R$ , so

X = \left(\frac{l_x}{l_R}\right)R

with R known and $l_x, l_R$ read directly off the metre scale. As described in Section 9.3.1, accuracy is improved by choosing R so the null point falls in the wire's MIDDLE THIRD, repeating the measurement with X and R swapped between the two gaps, and always tapping (never sliding) the jockey. [!ANSWER] $X = (l_x/l_R)\,R$ , found from the balance-point lengths on either side of the jockey.

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2026Set ANNUAL2 marks
$Q. State any two sources of error in meter bridge experiment. Explain how they can be minimised.$
›Reveal solutionSolution

The two main sources of error in a meter bridge are end resistance at the copper strips and non-uniformity of the bridge wire; both are corrected by calibration/measurement technique rather than by circuit change.

(1) End resistance (end correction): The resistance of the copper strips at the two ends of the meter bridge wire, and of the connections/jockey contact, adds an unknown extra resistance not accounted for by the measured length alone, causing a small error in the balance-point reading.

Minimisation: Determine the end correction experimentally (by interchanging known resistances in the two gaps and comparing the calculated vs. actual balance lengths), and add this correction to every observed balancing length; alternatively, keep the unknown and known resistances comparable so the balance point falls near the middle of the wire, where end effects influence the percentage error least.

(2) Non-uniform cross-section / material of the wire: If the meter-bridge wire does not have a strictly uniform cross-sectional area or uniform resistivity along its length, the assumption $R \propto l$ (resistance proportional to length) breaks down, introducing error.

Minimisation: Use a wire of uniform diameter (checked by a screw gauge at several points) made of a single, homogeneous material (e.g. constantan/manganin), and take the mean of several observations obtained by interchanging the positions of the known and unknown resistances.

✓Final answer
$(1) End resistance of the copper strips/connections — minimised by applying an end correction. (2) Non-uniform cross-section of the bridge wire — minimised by using a wire of uniform, carefully calibrated cross-section and taking readings near the middle of the wire.$
CBSE 2017Set ANNUAL2 marks
$Q. An unknown resistance is placed in the left gap and resistance of 50 ohm is placed in the right gap of a meter bridge. The null point is obtained at 40 cm from the left end. Determine the unknown resistance.$
›Reveal solutionSolution

Meter bridge balance condition: $\dfrac{X}{R} = \dfrac{l}{100-l}$ .

In a meter bridge, with unknown resistance $X$ in the left gap and known resistance $R$ in the right gap, if the null (balance) point is at length $l$ (cm) from the left end, the balance (Wheatstone bridge) condition gives

$\frac{X}{R} = \frac{l}{100-l}$

Given $R = 50\ \Omega$ , $l = 40\ \text{cm}$ :

$X = R\times\frac{l}{100-l} = 50\times\frac{40}{60} = 50\times\frac{2}{3} = 33.33\ \Omega$

✓Final answer
$X \approx 33.33\ \Omega$

$$X = \left(\frac{l_x}{l_R}\right)R$$

with R known and $l_x, l_R$ read directly off the metre scale. As described in Section 9.3.1, accuracy is improved by choosing R so the null point falls in the wire's MIDDLE THIRD, repeating the measurement with X and R swapped between the two gaps, and always tapping (never sliding) the jockey. [!ANSWER] $X = (l_x/l_R)\,R$, found from the balance-point lengths on either side of the jockey.

Q.Explain with a neat circuit diagram, how you will determine the unknown resistance by using a meter-bridge.

Concept understanding — Metre Bridge

Q.Explain with a neat circuit diagram, how you will determine the unknown resistance by using a meter-bridge.

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