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Chapter 26: Problem 35

Consider the potentiometer circuit of Fig. 26.19 \(\mathrm{a}\) . The resistor between \(a\) and \(b\) is a uniform wire with length \(l,\) with a sliding contact \(c\) at a distance \(x\) from \(b\) . An unknown \(\mathrm{emf} \mathcal{E}_{2}\) is measured by \(\mathcal{E}_{2}=(x / l) \mathcal{E}_{1} .\) (b) Why is the intermal resistance of the galvanometer not important? (c) Suppose \(\mathcal{E}_{1}=9.15 \mathrm{V}\) and \(l=1.000 \mathrm{m} .\) The galvanometer \(G\) reads zero when \(x=0.365 \mathrm{m} .\) What is the emf \(\mathcal{E}_{2} ?\)

Short Answer

Expert verified
\( \mathcal{E}_2 = 3.34 \text{ V} \).

Step by step solution

01

Understand the formula

The formula for the unknown emf \( \mathcal{E}_2 \) given the length \( x \) is \( \mathcal{E}_2 = \left( \frac{x}{l} \right) \mathcal{E}_1 \). This formula is based on the ratio of the lengths being proportional to the emfs when the circuit is balanced (galvanometer reads zero).
02

Substitute known values

We know \( \mathcal{E}_1 = 9.15 \text{ V} \), \( l = 1.000 \text{ m} \), and \( x = 0.365 \text{ m} \). Substitute these values into the formula to get \[ \mathcal{E}_2 = \left( \frac{0.365}{1.000} \right) \times 9.15 \].
03

Calculate \( \frac{x}{l} \)

Perform the division \( \frac{0.365}{1.000} \) to get \( 0.365 \).
04

Calculate \( \mathcal{E}_2 \) using the proportional value

Multiply the result from Step 3 by \( \mathcal{E}_1 \): \( 0.365 \times 9.15 = 3.34075 \).
05

Analyze why the galvanometer's internal resistance is irrelevant

When the galvanometer reads zero, it indicates a balanced condition in the potentiometer circuit. In this state, no current flows through the galvanometer, making its internal resistance irrelevant to the measurement.

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Key Concepts

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

Balanced Circuit
In a potentiometer circuit, a balanced condition is achieved when the pointer of the galvanometer reads zero. This means that there is no potential difference across the galvanometer and no current flows through it. Achieving this balance involves adjusting a sliding contact along a uniform wire until the electromotive force (emf) of the unknown source matches the proportional part of the known voltage drop along the wire.

The balance ensures that the ratio of the segment lengths of the wire between the terminals matches the ratio of the emfs. This forms the basis of potentiometer measurements and allows for precise voltage evaluation.
  • Helps measure unknown emf without drawing current from the circuit.
  • Ensures accuracy and eliminates potential errors due to current.
  • The division of the wire provides a method of scaling known voltage to match unknown values safely.
Electromotive Force Measurement
The potentiometer measures the electromotive force (emf) by using a principle of potential difference scaling along a wire. The known voltage source across the length of the wire creates a linear scale represented by the length of the wire.

In this setup, the unknown emf is compared to fractions of the known emf by adjusting the contact point on the wire. When balance is obtained, the unknown emf \(\mathcal{E}_2\) can be calculated with the formula: \[ \mathcal{E}_2 = \left( \frac{x}{l} \right) \mathcal{E}_1 \]
Here, \(x\) is the length of the wire corresponding to the unknown emf, \(l\) is the total length of the uniform wire, and \(\mathcal{E}_1\) is the known emf. This calculation method ensures that the measured emf is accurate as long as the wire and the initial calibration are precise.
Galvanometer in Circuit Analysis
In the potentiometer setup, the galvanometer plays a crucial role in indicating balance by showing zero deflection when there is no potential difference across it. This lack of deflection denotes a balanced circuit where the unknown emf equals the known emf proportionally scaled by the length of the wire.

Importantly, while balance is achieved, the internal resistance of the galvanometer no longer affects the circuit because no current flows through it. This detail simplifies the measurement process since adjustments or compensations for galvanometer resistance are unnecessary.
  • Acts as a null indicator in balanced circuits.
  • Ineffectual internal resistance ensures straightforward calculations.
  • Critical for determining exact points of equilibrium without needing current-drawing adjustments.
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Most popular questions from this chapter

A circuit consists of a series combination of \(6.00-\mathrm{k} \Omega\) and 5.00- \(\mathrm{k} \Omega\) resistors connected across a \(50.0-\mathrm{V}\) battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the \(5.00-\mathrm{k} \Omega\) resistor using a voltmeter having an internal resistance of 10.0 \(\mathrm{k} \Omega\) (a) What potential difference does the voltmeter measure across the \(5.00-\mathrm{k} \Omega\) resistor? (b) What is the true potential difference across this resistor when the meter is not present? (c) By what percentage is the voltmeter reading in error from the true potential difference?

A \(4.60-\mu F\) capacitor that is initially uncharged is connected in series with a \(7.50-\mathrm{k} \Omega\) resistor and an emf source with \(\mathcal{E}=125 \mathrm{V}\) and negligible internal resistance. Just after the circuit is completed, what are (a) the voltage drop across the capacitor; (b) the voltage drop across the resistor; (c) the charge on the capacitor; (d) the current through the resistor? (e) A long time after the circuit is completed (after many time constants) what are the val- ues of the quantities in parts (a)-(d)?

Three identical resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 27 \(\mathrm{W}\) . What power would be dissipated if the three resistors were connected in parallel across the same potential difference?

A certain galvanometer has a resistance of 65.0\(\Omega\) and deflects full scale with a current of 1.50 \(\mathrm{mA}\) in its coil. This is to be replaced with a second galvanometer that has a resistance of 38.0\(\Omega\) and deflects full scale with a current of 3.60\(\mu A\) in its coil. Devise a circuit incorporating the second galvanometer such that the equivalent resistance of the circuit equals the resistance of the first galvanometer, and the second galvanometer deflects full scale when the current through the circuit equals the full-scale current of the first galvanometer.

Two identical \(1.00-\Omega\) wires are laid side by side and soldered together so they touch each other for half of their lengths. What is the equivalent resistance of this combination?
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