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Chapter 18: Problem 61

\(A, B, C\) and \(D\) are four resistances of \(2 \Omega, 2 \Omega, 2 \Omega\) and \(3 \Omega\) respectively. They are used to form a Wheatstone bridge. The resistance \(D\) is short circuited with a resistance \(R\) in order to get the bridge balanced. The value of \(R\) will be (a) \(4 \Omega\) (b) \(6 \Omega\) (c) \(8 \Omega\) (d) \(3 \Omega\)

Short Answer

Expert verified
The value of \( R \) is \( 2 \Omega \).

Step by step solution

01

Understand the Wheatstone Bridge

In a Wheatstone bridge setup, the bridge is balanced when the ratio of resistances in one branch is equal to the ratio in the other branch. The bridge has four resistances: two in each of the two branches connected in series.
02

Assign the resistances

The given values for the resistances are: \( A = 2 \Omega \), \( B = 2 \Omega \), \( C = 2 \Omega \), and \( D = 3 \Omega \). Resistance \( D \) is short-circuited with another resistance \( R \) to form a balanced bridge.
03

Set up the ratio equation for a balanced condition

In a balanced condition, the ratio \( \frac{A}{B} = \frac{C}{R} \) must hold true. Substitute the known values: \( \frac{2 \Omega}{2 \Omega} = \frac{2 \Omega}{R} \).
04

Solve for R

Since \( \frac{2}{2} = \frac{2}{R} \), this simplifies to \( 1 = \frac{2}{R} \). Multiply both sides by \( R \) to get \( R = 2 \Omega \).

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Key Concepts

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

Balanced Condition in a Wheatstone Bridge
A critical concept behind the Wheatstone bridge is its balanced condition. This occurs when the ratio of resistances in one path equals the ratio in the other path of the bridge. The beauty of this condition is that no current flows through the bridge's measuring instrument, often a galvanometer. This unique balance allows for accurate measurement of an unknown resistance.
In our case, resistances A and B form one path, while C and D form another. For the bridge to be balanced, the condition \( \frac{A}{B} = \frac{C}{R} \) must be satisfied. This means that any change, like the introduction of a new resistance, needs reevaluation to maintain equilibrium. Understanding this ratio-based balance is key to utilizing the Wheatstone bridge effectively in electrical circuits.
Resistance Calculation for Balance
Calculating resistance to maintain balance in a Wheatstone bridge involves simple algebraic manipulation. First, you frame the balance condition in the form of \( \frac{A}{B} = \frac{C}{R} \).
In the given problem, all resistances except \( R \) are known. By plugging these values into the equation \( \frac{2 \Omega}{2 \Omega} = \frac{2 \Omega}{R} \), it simplifies to \( 1 = \frac{2}{R} \).

To solve for \( R \), multiply both sides by \( R \), giving \( R = 2 \Omega \). This calculation ensures the bridge is balanced, highlighting how straightforward mathematical proportions can solve practical engineering problems.
Understanding Electrical Circuits and Wheatstone Bridge
Electrical circuits are composed of paths that electricity flows through, including components like resistors, which impede flow, forcing a voltage drop. The Wheatstone bridge, an essential part of these circuits, helps measure an unknown resistance accurately.
It operates with four resistors: two sets in series, forming two parallel branches. One "bridge" joins the branches at midpoints, completing the circuit. This bridge carries the measuring device.
  • When balanced, accurate resistance values are determined without current disturbing measurements.
  • This setup is helpful for precision tasks, such as calibrating instruments.
  • The bridge highlights concepts such as symmetry in circuits and the non-linear behavior of resistors in series or parallel.
Grasping these fundamentals of how circuits and bridges interplay paves the way for more complex electronics learning.

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Most popular questions from this chapter

The temperature of the cold junction of a thermocouple is \(0{ }^{\circ} \mathrm{C}\) and the temperature of hot iunction is \(T^{\circ} \mathrm{C}\). The emf is \(E=16 T-0.04 T^{2} \mu \mathrm{V}\). The inversion temperature \(T_{i}\) is (a) \(200^{\circ} \mathrm{C}\) (b) \(400^{\circ} \mathrm{C}\) (c) \(100^{\circ} \mathrm{C}\) (d) \(300^{\circ} \mathrm{C}\)

In a neon gas discharge tube \(\mathrm{Ne}^{+}\)ions moving through a cross- section of the tube each second to the right is \(2.9 \times 10^{18}\), while \(1.2 \times 10^{18}\) electrons move towards left in the same time. The electronic charge being \(1.6 \times 10^{-19} \mathrm{C}\), the net electric current is (a) \(0.27\) A to the right (b) \(0.66\) A to the right (c) \(0.66 \mathrm{~A}\) to the left (d) zero

An electric heater of \(1.08 \mathrm{~kW}\) is immersed in water. After the water has reached a temperature of \(100^{\circ} \mathrm{C}\), how much time will be required to produce \(100 \mathrm{~g}\) of steam? (Latent heat of steam \(=540 \mathrm{calg}^{-1}\) )

Assertion The temperature depending on resistance is usually given as \(R=R_{0}(1+\alpha \Delta t)\). The resistance of a wire changes from \(100 \Omega\) to \(150 \Omega\). When its temperature is increased from \(27^{\circ} \mathrm{C}\) to \(227^{\circ} \mathrm{C}\). This implies that \(\alpha=2.5 \times 10^{-3}{ }^{\circ} \mathrm{C}^{-1}\). Reason \(R=R_{0}(1+\alpha \Delta t)\) is valid when then change in the temperature \(\Delta T\) is small and \(\Delta R=\left(R-R_{0}\right)<

A battery of emf \(E\) and internal resistance \(r\) is connected with an external voltage source (generator) through a resistance \(R\) as shown in figure. Choose the correct statements. (a) In order to charge the battery, the output voltage \(V\) of the generator must be greater than \(E\) (b) In order to charge the battery, the output voltage \(V\) of the generator must be at least twice of \(E\) (c) The charging current \(i\) through the circuit is given by \(i=\frac{V-E}{(R+r)}\) (d) The charging current \(i\) through the circuit is given by \(i=\frac{V}{(R+r)}\)
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