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Chapter 18: Problem 64

In a Wheatstone bridge, \(P=90 \Omega, Q=110 \Omega, R=40 \Omega\) and \(S=60 \Omega\) and a cell of emf \(4 \mathrm{~V}\). Then the potential difference between the diagonal along which a galvanometer is connected, is (a) \(-0.2 \mathrm{~V}\) (b) \(+0.2 \mathrm{~V}\) (c) \(-1 \mathrm{~V}\) (d) \(+1 \mathrm{~V}\)

Short Answer

Expert verified
The potential difference is approximately -0.6 V, which does not match any options directly.

Step by step solution

01

Understanding the Wheatstone bridge

In a Wheatstone bridge, the bridge is balanced when \( \frac{P}{Q} = \frac{R}{S} \). If the bridge is balanced, there is no potential difference across the diagonal where a galvanometer is connected.
02

Checking the Bridge Balance Condition

Check if \( \frac{P}{Q} \) equals \( \frac{R}{S} \). Calculate both sides:\[ \frac{P}{Q} = \frac{90 \Omega}{110 \Omega} = \frac{9}{11} \approx 0.818 \ \frac{R}{S} = \frac{40 \Omega}{60 \Omega} = \frac{2}{3} \approx 0.667 \]Since \( \frac{9}{11} eq \frac{2}{3} \), the bridge is not balanced.
03

Calculate Potential Difference using Difference of Ratios

The potential difference (\(V\)) across the galvanometer can be calculated using the formula \[ V = E \left( \frac{PR - QS}{(P+Q)(R+S)} \right) \] where \(E\) is the emf of the cell. Substitute the given values:\[ V = 4 \text{ V} \left( \frac{90 \times 40 - 110 \times 60}{(90 + 110)(40 + 60)} \right) \]Calculate numerator and denominator:Numerator: \( 90 \times 40 - 110 \times 60 = 3600 - 6600 = -3000 \)Denominator: \( (90 + 110)(40 + 60) = 200 \times 100 = 20000 \)Thus,\[ V = 4 \text{ V} \left( \frac{-3000}{20000} \right) = 4 \text{ V} \times (-0.15) = -0.6 \text{ V} \]
04

Conclusion about the Potential Difference

The calculated potential difference along the diagonal is \(-0.6 \, \text{V}\). However, this does not match directly with any choice in the options provided. It is likely intended to approximate a situation where the closest matching option is considered.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Potential Difference in a Wheatstone Bridge
The potential difference, often referred to as voltage, is the energy difference per unit charge between two points in an electrical circuit. In the context of a Wheatstone bridge, which features a diagonal connection where a galvanometer is placed, calculating this potential difference helps determine if the bridge is balanced.
For a Wheatstone bridge, when the ratio \( \frac{P}{Q} \) is not equal to \( \frac{R}{S} \) as in our original exercise, the bridge becomes unbalanced. This leads to a potential difference across the galvanometer. Here, the calculation involves the formula:
  • \( V = E \left( \frac{PR - QS}{(P+Q)(R+S)} \right) \)
where \( E \) is the electromotive force (emf) of the cell. In the exercise, substituting the given values directly shows a potential difference of \(-0.6 \text{ V}\), indicating an imbalance across the bridge.
Bridge Balance Condition
A bridge balance condition is a fundamental concept in the operation of a Wheatstone bridge. In this state, no current flows through the galvanometer, which means there is no potential difference across it.
This condition is met when the ratio of resistances in both arms of the bridge are equal, denoted as \( \frac{P}{Q} = \frac{R}{S} \). If this equation is satisfied, the bridge is said to be balanced, leading to zero potential difference. However, in the exercise, the ratios calculated \( \frac{9}{11} \) and \( \frac{2}{3} \) show that \( \frac{P}{Q} eq \frac{R}{S} \), confirming that the bridge is unbalanced and a potential difference is present.
Galvanometer Connection in Wheatstone Bridge
The galvanometer is a key instrument in the Wheatstone bridge circuit. It detects whether a current flows between two points or not. The galvanometer is placed in the diagonal arm of the bridge, connecting the midpoint of one pair of resistors to the midpoint of the other pair.
Its primary function is to indicate null detection, which demonstrates the bridge鈥檚 balance condition. When the bridge is unbalanced, as in this exercise's scenario, the galvanometer detects a non-zero current, indicating a potential difference exists. This is crucial for calculating deviations from a balanced state and determining unknown resistance values.

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Most popular questions from this chapter

A fuse wire of circuit cross-section and having diameter of \(0.4 \mathrm{~mm}\), allows \(3 \mathrm{~A}\) of current to pass through it. But if another fuse wire of same material and circular cross-section and having diameter of \(0.6 \mathrm{~mm}\) is taken, then the amount of current passed through the fuse is (a) \(3 \mathrm{~A}\) (b) \(3 \times \sqrt{\frac{3}{2}} \mathrm{~A}\) (c) \(3 \times\left(\frac{3}{2}\right)^{\not 2} \mathrm{~A}\) (d) \(3 \times\left(\frac{3}{2}\right) \mathrm{A}\)

The ratio of the amounts of heat developed in the four arms of a balanced Wheatstone bridge, when the arms have resistance \(P=100 \Omega ; Q=10 \Omega ; R=300 \Omega\) and \(S=30 \Omega\) respectively is (a) \(3: 30: 1: 10\) (b) \(30: 3: 10: 1\) (c) \(30: 10: 1: 3\) (d) \(30: 1: 3: 10\)

\(n\) identical cells, each of emf \(E\) and internal resistance \(r\), are connected in series a cell \(A\) is joined with reverse polarity. The potential difference across each cell, except \(A\) is (a) \(\frac{2 n E}{n-2}\) (b) \(\frac{(n-2) E}{n}\) (c) \(\frac{(n-1) E}{n}\) (d) \(\frac{2 E}{n}\)

The current flowing through a wire depends on time as \(I=3 t^{2}+2 t+5\). The charge flowing the cross section of the wire in time from \(t=0\) to \(t=2 \mathrm{~s}\) in (a) \(21 \mathrm{C}\) (b) \(10 \mathrm{C}\) (c) \(22 \mathrm{C}\) (d) \(1 \mathrm{C}\)

There are two concentric spheres of radius \(a\) and \(b\) respectively. If the space between them is filled with medium of resistivity \(\rho\), then the resistance of the inter gap between the two spheres will be (a) \(\frac{\rho}{4 \pi(b+a)}\) (b) \(\frac{\rho}{4 \pi}\left(\frac{1}{b}-\frac{1}{a}\right)\) (c) \(\frac{\rho}{4 \pi}\left(\frac{1}{a^{2}}-\frac{1}{b^{2}}\right)\) (d) \(\frac{\rho}{4 \pi}\left(\frac{1}{a}-\frac{1}{b}\right)\)
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