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Current Electricity
Meter Bridge
2 questions from 2 papers, each with its answer.
- questions
- 2
- papers
- 2
- years asked, 2017–2026
- 1 of 10
- of Current Electricity
- 1.54%
Appears in
Concepts tested
- Balance condition
- resistance per unit length from a metre bridge
All Questions
2020 1 question
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A resistance wire connected in the left gap of a metre bridge balances a $10\ \Omega$ resistance in the right gap at a point which divides the bridge wire in the ratio $3:2$. If the length of the resistance wire is 1.5 m, then the length of $1\ \Omega$ of the resistance wire is:
At balance: $\frac{P}{10} = \frac{l_1}{l_2} = \frac{3}{2} \Rightarrow P = 15\ \Omega$
The resistance wire of length 1.5 m has resistance 15 Ω, and $R \propto l$:
$\frac{15}{1} = \frac{1.5}{l_2} \Rightarrow l_2 = 0.1$ m $= 1.0 \times 10^{-1}$ m
2014 AIPMT 1 question
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The resistances in the two arms of the meter bridge are 5 $\Omega$ and R $\Omega$, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at $1.6l_1$. The resistance 'R' is:
$\frac{5}{R} = \frac{l_1}{100 - l_1}$ and $\frac{5}{R/2} = \frac{1.6l_1}{100 - 1.6l_1}$
Solving, $l_1 = 25$ cm and $R = 15\ \Omega$