Question

To understand the units of inductance, the potential energy stored in an inductor, and some of the consequences of having inductance in a circuit.

After batteries, resistors, and capacitors, the most common elements in circuits are inductors. Inductors usually look like tightly wound coils of fine wire. Unlike capacitors, which produce a physical break in the circuit between the capacitor plates, the wire of an inductor provides an unbroken continuous path in which current can flow. When the current in a circuit is constant, an inductor acts essentially like a short circuit (i.e., a zero-resistance path). In reality, there is always at least a small amount of resistance in the windings of an inductor, a fact that is usually neglected in introductory discussions.

Recall that current flowing through a wire generates a magnetic field in the vicinity of the wire. If the wire is coiled , such as in a solenoid or an inductor, the magnetic field is strongest within the coil parallel to its axis. The magnetic field associated with current flowing through an inductor takes time to create, and time to eliminate when the current is turned off. When the current changes, an EMF is generated in the inductor, according to Faraday's law, that opposes the change in current flow. Thus inductors provide electrical inertia to a circuit by reducing the rapidity of change in the current flow.

Figure 1 of 1 Graph A 3.5 E 2.5 g 1.5 1 0 0.5 1 1.5 2 Time (ms) Graph C 3.5 2.5 E 1.5 C 1 0.5 0 0.5 1 1.5 2 Time (ms) Graph B 3.5 E 2.5 g 1.5 0.5 0 0.5 1 1.5 2 Time (ms) Graph D 3.5 E 2.5 1.5 1 0.5 0 0.5 1 1.5 2 Time ms)

Answer 1

PART A

It is given

ε = -LΔI/Δt

-εΔt/ΔI = L

Hence the unit is Vs/A

1H = 1(V.s/A)

PART B

The inductance of the waffle iron L = 560 x 10^-3 H = 0.56 H

Initial current flowing in circuit I = 2.5 A

Tame taken for current to become zero Δt = 0.015 s

ε = -LΔI/Δt

ε = - 0.56 x (0 - 2.5)/0.015

ε = 93.33 V

PART C

1. Increasing the inductance by a factor of 5; leaving the current unchanged
2. Leaving the inductance unchanged; increasing the current by a factor of √5
3. Reducing the inductance by a factor of 5; increasing the current by a factor of 5

PART D

As per the given information, the curve of Current Vs time must be continuous and the slope must not be infinite

Hence the curves A, B and C are possible graphs

Answer is ABC

Answer 2

part A

since,

$\varepsilon =- L \,\frac{\Delta I}{\Delta t}$

by re-arranging, we can write

$L = -\varepsilon \,\frac{\Delta t}{\Delta I}$

now, unit of inductor is Henry, unit of time is second and unit of current and emf is A and V, therefore

$1H = V \,\frac{s}{A}$

Part -B

Given that,

The inductance of the waffle iron is,

$L = 560mH= 560 \times 10^{-3}H$

and Initial current flowing in circuit is ,

and time taken by current to become zero is,

$\Delta t = 0.015 s$

therefore,

$\varepsilon =- L \,\frac{\Delta I}{\Delta t}$

$\varepsilon =-(0.56H) \,\left( \frac{0-2.5A}{0.015s} \right)$

$\varepsilon = 93.33 V$

part C

1. Increasing the inductance by a factor of 5; leaving the current unchanged

$U = \frac{1}{2}(5L)I^2 \Rightarrow 5U$

2. Leaving the inductance unchanged; increasing the current by a factor of √5

   $U = \frac{1}{2}L(\sqrt{5}I)^2 = \frac{1}{2}L(5I^2)\Rightarrow 5U$

3. Reducing the inductance by a factor of 5; increasing the current by a factor of 5,

   $U = \frac{1}{2}\left(\frac{L}{5}\right)(5I)^2 = \frac{1}{2}\left(\frac{L}{5} \right)(25I^2)\Rightarrow 5U$

Part d

As per the given information, the curve of Current Vs time must be continuous and the slope must not be infinite

Hence the curves A, B and C are possible graphs

Answer is ABC