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Chapter 16: Problem 5

SI unit of inductance can be written as (A) Weber/Ampere (B) Joule/Ampere \(^{2}\) (C) Ohm/Second (D) All of the above

Short Answer

Expert verified
The correct SI unit of inductance is (A) Weber/Ampere, which can be represented as \(1 \, \text{Henry} = 1 \, \dfrac{\text{Weber}}{\text{Ampere}}\).

Step by step solution

01

Recall the definition of inductance

Inductance is a property of an electrical circuit that opposes any change in current. It depends on the physical characteristics of a conductor and the magnetic field around it. The symbol for inductance is \(L\), and its SI unit is called the Henry (H).
02

Express the SI unit of inductance in terms of basic units

Recall that the definition of Henry can be written in terms of other SI units: \[1 \, \text{Henry} = 1 \, \dfrac{\text{Weber} \cdot \text{Volt}}{\text{Ampere}^2}\] We also know that Volt/Amper = Ohm and Weber = Volt * second. So, the expression becomes: \[1 \, \text{Henry} = 1 \, \dfrac{(\text{Volt} \cdot \text{Second}) \cdot \text{Ohm}}{\text{Ampere}^2}\]
03

Simplify the expression and compare with the given options

The simplified expression for the SI unit of inductance is: \[1 \, \text{Henry} = 1 \, \dfrac{\text{Weber}}{\text{Ampere}}\] Now, let's compare this expression with the given options: (A) Weber/Ampere: This matches our simplified expression, so this option is correct. (B) Joule/Ampere虏: This option does not match our simplified expression, so this option is incorrect. (C) Ohm/Second: This option does not match our simplified expression, so this option is incorrect. (D) All of the above: Since only option (A) is correct, this option is also incorrect.
04

Identify the correct answer

The correct answer is (A) Weber/Ampere. This is the correct SI unit of inductance.

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Most popular questions from this chapter

The diagram shows a solenoid carrying time varying current \(I=I_{0} t\). On the axis of this solenoid, a ring has been placed. The mutual inductance of the ring and the solenoid is \(M\) and the self-inductance of the ring is \(L\). If the resistance of the ring is \(R\) then maximum current which can flow through the ring is (A) \(\frac{(2 M+L) I_{0}}{R}\) (B) \(\frac{M I_{0}}{R}\) (C) \(\frac{(2 M-L) I_{0}}{R}\) (D) \(\frac{(M+L) I_{0}}{R}\)

If the flux of magnetic induction through a coil of resistance \(R\) and having \(n\) turns changes from \(\phi_{1}\) to \(\phi_{2}\), then the magnitude of the charge that passes through the coil is (A) \(\frac{\left(\phi_{2}-\phi_{1}\right)}{R}\) (B) \(\frac{n\left(\phi_{2}-\phi_{1}\right)}{R}\) (C) \(\frac{\left(\phi_{2}-\phi_{1}\right)}{n R}\) (D) \(\frac{n R}{\left(\phi_{2}-\phi_{1}\right)}\)

Flux \(\phi\) (in weber) in a closed circuit of resistance \(10 \Omega\) varies with time \(t\) (in seconds) according to the equation \(\phi=6 t^{2}-5 t+1\). The magnitude of the induced current in the circuit at \(t=0.25 \mathrm{~s}\) is (A) \(0.2 \mathrm{~A}\) (B) \(0.6 \mathrm{~A}\) (C) \(0.8 \mathrm{~A}\) (D) \(1.2 \mathrm{~A}\)

An iron rod of cross-sectional area 4 sq \(\mathrm{cm}\) is placed with its length parallel to a magnetic field of intensity \(1600 \mathrm{amp} / \mathrm{m}\). The flux through the rod is \(4 \times 10^{-4}\) weber. The permeability of the material of the rod is (In weber/amp-m). (A) \(0.625\) (B) \(6.25\) (C) \(0.625 \times 10^{-3}\) (D) None of these

A circular loop of radius \(0.3 \mathrm{~cm}\) lies parallel to a much bigger circular loop of radius \(20 \mathrm{~cm}\). The centre of the small loop is on the axis of the bigger loop. The distance between their centres is \(15 \mathrm{~cm}\). If a current of \(2.0 \mathrm{~A}\) flows through the smaller loop, then the flux linked with bigger loop is (A) \(6 \times 10^{-11}\) weber (B) \(3.3 \times 10^{-11}\) weber (C) \(6.6 \times 10^{-9}\) weber (D) \(9.1 \times 10^{-11}\) weber
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