/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 An ammeter reads up to \(1 \math... [FREE SOLUTION] | 91影视

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Chapter 14: Problem 33

An ammeter reads up to \(1 \mathrm{~A}\). Its internal resistance is \(0.81 \Omega\). To increase the range to \(10 \mathrm{~A}\), the value of the required shunt is (A) \(0.03 \Omega\) (B) \(0.3 \Omega\) (C) \(0.9 \Omega\) (D) \(0.09 \Omega\)

Short Answer

Expert verified
The shunt resistance required to increase the range of the ammeter to 10 Amperes is approximately \(0.09 \, \Omega\).

Step by step solution

01

Calculate the Ammeter's Maximum Voltage Drop

First, find the maximum voltage drop across the ammeter when it reads its full-scale deflection of 1 Ampere. Use Ohm's law for this, which is \( V = I \times R \), where \( V \) is the voltage, \( I \) is the current, and \( R \) is the resistance. The voltage across the ammeter is \( V = 1A \times 0.81 \, \Omega = 0.81 \, volts \).
02

Calculate the Total Current Required

Second, since the ammeter needs to be able to read up to 10 Amperes, the total current (including the initial 1 Ampere the ammeter can already measure) is 10 Amperes.
03

Determine the Current Across the Shunt Resistor

Third, determine the current that needs to pass through the shunt resistor. This is the total current minus the current the ammeter already handles (so \( I_{shunt} = I_{total} - I_{ammeter} \)). The shunt current is \( I_{shunt} = 10A - 1A = 9A \).
04

Calculate the Shunt Resistance

Finally, determine the shunt resistance using Ohm's law again. Rearrange the equation to solve for the resistance: \( R = \frac{V}{I} \). The shunt resistance is \( R_{shunt} = \frac{0.81 \, volts}{9A} \approx 0.09 \, \Omega \). Therefore, the correct option is (D) \(0.09 \Omega\).

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Most popular questions from this chapter

This question has statement I and statement II. Of the four choices given after the statements, choose the one that best describes the two statements. Statement-I: Higher the range, greater is the resistance of ammeter. Statement-II: To increase the range of ammeter, additional shunt needs to be used across it. (A) Statement-I is true, Statement-II is true, Statement-II is not the correct explanation of Statement-I. (B) Statement-I is true, Statement-II is false. (C) Statement-I is false, Statement-II is true. (D) Statement-I is true, Statement-II is true, Statement-II is correct explanation of Statement-I.

The resistance of the series combination of two resistances is \(S\). When they are joined in parallel, the total resistance is \(P\). If \(S=n P\), then the minimum possible value of \(n\) is (A) 4 (B) 3 (C) 2 (D) 1

In the given circuit, find the equivalent resistance between points \(A\) and \(B\). (A) \(18 \Omega\) (B) \(12 \Omega\) (C) \(20 \Omega\) (D) \(27 \Omega\)

A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be (Assuming potential difference is same in both cases). (A) One-fourth (B) Halved (C) Doubled (D) Four times

Figure \(14.47\) shows the circuit of a potentiometer. The length of the potentiometer wire \(A B\) is \(50 \mathrm{~cm}\). The EMF \(E_{1}\) of the battery is \(4 \mathrm{~V}\), having negligible internal resistance. Value of \(R_{1}\) and \(R_{2}\) are \(15 \Omega\) and \(5 \Omega\), respectively. When both the keys are open, the null point is obtained at a distance of \(31.25 \mathrm{~cm}\) from \(A\), but when both the keys are closed, the balance length reduces to \(5 \mathrm{~cm}\) only. Given \(R_{A B}=10 \Omega\) The balance length when key \(K_{2}\) is open and \(K_{1}\) is closed is given by (A) \(10.5 \mathrm{~cm}\) (B) \(11.5 \mathrm{~cm}\) (C) \(12.5 \mathrm{~cm}\) (D) \(13.5 \mathrm{~cm}\)
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