Cells, Kirchhoff’s Laws, Bridges and Potentiometer — Part 2: Bridges and the potentiometer: visual lesson, ISC Class 12 Physics

Lesson

Bridges and the potentiometer

Part 2 — null methods: instruments that measure by finding where the current is zero.

In this lesson you will

Derive the balance condition of a Wheatstone bridge and use a metre bridge to find an unknown resistance
Explain the principle of the potentiometer, its potential gradient and how to make it more sensitive
Use a potentiometer to compare emfs and to find the internal resistance of a cell

Worked example 6: k = 0.40 V m⁻¹, balance at 2.7 m.

Diagram

The Wheatstone bridge

Four resistors P, Q, R, S, a cell across AC and a galvanometer across BD. At balance ( $I_g = 0$ ), B and D are at the same potential, so the same current $I_1$ flows through P and Q and $I_2$ through R and S: $I_1P = I_2R,\quad I_1Q = I_2S\ \Rightarrow\ \dfrac{P}{Q} = \dfrac{R}{S}$

Loop ABDA with $I_g = 0$ gives the first equation; loop BCDB the second.
Balance does not depend on the emf of the cell or the resistance of the galvanometer.

Diagram

The metre bridge

Worked example 5: balance at 55.0 cm.

A practical Wheatstone bridge: a uniform wire 100 cm long forms the ratio arms. With X in the left gap and a known R in the right, balance at length l gives $\dfrac{X}{R} = \dfrac{l}{100 - l}\qquad\Rightarrow\qquad X = R\,\dfrac{l}{100 - l}$

The wire’s resistance is proportional to length, so the arms are in the ratio l : (100 − l).
Choose R so that l is near the middle: the end resistances at the copper strips then matter least.
Touch the jockey gently; do not slide it along the wire.

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