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Meter Bridge: Unknown Resistance

Lesson 4 of 8 3D virtual lab schedule15 min

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flagWhat you'll discover

  • arrow_forwardExplain the Wheatstone bridge balance condition
  • arrow_forwardLocate the null point with a jockey on the bridge wire
  • arrow_forwardCalculate X = R(100 − l)/l from the balancing length
  • arrow_forwardUse the interchange method to cancel end errors

A Wheatstone bridge made of wire

The meter bridge is a folded Wheatstone bridge: known resistance R in the left gap, unknown X in the right gap, and a uniform 1 m constantan wire as the other two arms. A jockey tapped at distance l from the left end splits the wire into resistances proportional to l and (100 − l).

At the null point the galvanometer reads zero — no current crosses the bridge — and the balance condition gives R/X = l/(100 − l), so X = R(100 − l)/l. No meter calibration matters at balance: the galvanometer only needs to show ZERO honestly, which is why bridge methods beat ammeter-voltmeter methods for precision.

Procedure for a sharp null

Choose R from the resistance box so the balance lands near the middle of the wire (between 40 and 60 cm) — there the bridge is most sensitive and percentage errors in l are smallest. Tap the jockey briefly; never drag it, or you scrape the wire and change its cross-section.

Find the deflection direction at both ends first (it must reverse — otherwise a connection is broken), then close in on zero. Record l, then INTERCHANGE R and X and balance again: the average of X from both positions cancels the small "end resistances" where the wire is soldered to the copper strips. Repeat for two or three different R values.

Errors and standard viva questions

Sources of error: the wire may not be perfectly uniform (the biggest one), end/contact resistances at the copper strips, heating of the wire if current flows too long, and a blunt jockey contact. Precautions: use the key only while tapping, keep balance near 50 cm, interchange and average, and never press hard with the jockey.

Classic viva questions: "Why is the galvanometer not damaged at balance?" (no current flows through it), "Why constantan wire?" (uniform, high resistivity, low temperature coefficient), and "Why is the method unsuitable for very low or very high resistances?" (end resistances and insensitivity dominate when X is far from R).

quizCheck your knowledge

1. Balance occurs at l = 40 cm with R = 6 Ω in the left gap. X = ?
2. At the exact null point, the current through the galvanometer is:
3. Why should the balance point lie near the middle of the wire?