Numerical Problems-1 - Chapter 4 Electromagnetic Induction and Alternating Current 12th Science Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Numerical Problems
Question 1.
A square coil of side $30 \mathrm{~cm}$ with 500 turns is kept in a uniform magnetic field of $0.4 \mathrm{~T}$. The plane of the coil is inclined at an angle of $30^{\circ}$ to the field. Calculate the magnetic flux through the coil.
Solution:
Square coil of side $(a)=30 \mathrm{~cm}=30 \times 10^{-2} \mathrm{~m}$
Area of square coil $(\mathrm{A})=\mathrm{a}^2=\left(30 \times 10^{-2}\right) 2=9 \times 10^{-2} \mathrm{~m}^2$
Number of turns $(\mathrm{N})=500$
Magnetic field $(\mathrm{B})=0.4 \mathrm{~T}$
Angular between the field and coil $(\theta)=90-30=60^{\circ}$
Magnetic flux $(\Phi)=\mathrm{NBA} \cos 0=500 \times 0.4 \times 9 \times 10^{-2} \times \cos 60^{\circ}=18 \times \frac{1}{2}$
$\Phi=9 \mathrm{~Wb}$
Question 2.
A straight metal wire crosses a magnetic field of flux $4 \mathrm{mWb}$ in a time $0.4 \mathrm{~s}$. Find the magnitude of
the emf induced in the wire.
Solution:
Magnetic flux $(\Phi)=4 \mathrm{~m} \mathrm{~Wb}=4 \times 10^{-3} \mathrm{~Wb}$
time $(\mathrm{t})=0.4 \mathrm{~s}$
The magnitude of induced $\operatorname{emf}(\mathrm{e})=\frac{d \Phi}{d t}=\frac{4 \times 10^{-3}}{0.4} 10^{-2}$
$\mathrm{e}=10 \mathrm{mV}$

Question 3.
The magnetic flux passing through a coil perpendicular to its plane is a function of time and is given by $\mathrm{OB}=\left(2 t^3+4 t^2+8 t+8\right) \mathrm{Wb}$. If the resistance of the coil is $5 \Omega$, determine the induced current through the coil at a time $t=3$ second.
Solution:
Magnetic flux $\left(\Phi_{\mathrm{B}}\right)=\left(2 \mathrm{t}^3+8 \mathrm{t}^2+8 \mathrm{t}+8\right) \mathrm{Wb}$
Resistance of the coil $(\mathrm{R})=5 \Omega$ time $(t)=3$ second
Induced current through the coil, $\mathrm{I}=\frac{e}{R}$
Induced emf, e $=\frac{d \Phi_B}{d t}=\frac{d}{d t}\left(\left(2 \mathrm{t}^3+4 \mathrm{t}^2+8 \mathrm{t}+8\right)=6 \mathrm{t}^2+8 \mathrm{t}+8\right.$
Here time $(\mathrm{t})=3$ second
$\mathrm{e}=6(3)^2+8 \times 3+8=54+24+8=86 \mathrm{~V}$
$\therefore$ Induced current through the coil, $\mathrm{I}=\frac{e}{R}=\frac{86}{5}=17.2 \mathrm{~A}$
Question 4.
A closely wound coil of radius $0.02 \mathrm{~m}$ is placed perpendicular to the magnetic field. When the magnetic field is changed from $8000 \mathrm{~T}$ to $2000 \mathrm{~T}$ in $6 \mathrm{~s}$, an emf of $44 \mathrm{~V}$ is induced. Calculate the number of turns in the coil.
Solution:
Radius of the coil $(\mathrm{r})=0.02 \mathrm{~m}$
Area of the coil $(\mathrm{A})=\pi \mathrm{r}^2=3.14 \times(0.02)^2=1.256 \times 10^{-3} \mathrm{~m}^2$
Change in magnetic field, $\mathrm{dB}=8000-2000=6000 \mathrm{~T}$
Time, $\mathrm{dt}=6$ second
Induced emf, e $=44 \mathrm{~V}$
$\theta=0^{\circ}$
Induced emf in the coil, e $=\mathrm{NA} \frac{d}{d t} \cos \theta . \mathrm{dt}$
$44=\mathrm{N} \times 1.256 \times 10^{-3} \times \frac{600}{6} \times \operatorname{Cos} 0^{\circ}$
$\mathrm{N}=\frac{44}{1.256 \times 10^{-3} \times 1000 \times 1}=\frac{44}{1.256 \times 10^{-3} \times 10^3}$
Number of turns $\mathrm{N}=35$ turns
Question 5.
A rectangular coil of area $6 \mathrm{~cm}^2$ having 3500 turns is kept in a uniform magnetic field of $0.4 \mathrm{~T}$, Initially, the plane of the coil is perpendicular to the field and is then rotated through an angle of $180^{\circ}$. If the resistance of the coil is $35 \Omega$, find the amount of charge flowing through the coil.
Solution:
Rectangular coil of their area, $A=6 \mathrm{~cm}^2=6 \times 10^{-4} \mathrm{~m}^2$

Number of turns $N=3500$ turns
Magnetic field, $\mathrm{B}=0.4 \mathrm{~T}$
Resistance of the coil, $\mathrm{R}=35 \Omega$
Induced emf $(\mathrm{e})=$ change in flux per second $=\Phi_2-\Phi_1$
$\mathrm{e}=\mathrm{NAB} \cos 180^{\circ}-\mathrm{NBA} \cos 0^{\circ}=-\mathrm{NBA}-\mathrm{NBA}=-2 \mathrm{NBA}$
$=-2 \times 3500 \times 0.4 \times 6 \times 10^{-4}-16800 \times 10^{-4}=-1.68 \mathrm{~V}$
Current flowing the coil, $\mathrm{I}=\frac{e}{R}=\frac{-1.68}{35}=0.048$
Magnitude of the current, $I=48 \times 10^{-3} \mathrm{~A}$
Amount of charge flowing through the coil, $\mathrm{q}=\mathrm{It}=48 \times 10^{-3} \times 1=48 \times 10^{-3} \mathrm{C}$
Question 6.
An induced current of $2.5 \mathrm{~mA}$ flows through a single conductor of resistance $100 \Omega$. Find out the rate at which the magnetic flux is cut by the conductor.
Solution:
Induced current, $\mathrm{I}=2.5 \mathrm{~mA}$
Resistance of conductor, $\mathrm{R}=100 \Omega$
$\therefore$ The rate of change of flux, $\frac{d \Phi_B}{d t}=\mathrm{e}$
$\frac{d \Phi_B}{d t}=\mathrm{e}=\mathrm{IR}=2.5 \times 10-3 \times 100=250 \times 10^{-3} \mathrm{dt}$
$\frac{d \Phi_B}{d t}=250 \mathrm{mWb} \mathrm{s}^{-1}$
Question 7.
A fan of metal blades of length $0.4 \mathrm{~m}$ rotates normal to a magnetic field of $4 \times 10^{-3} \mathrm{~T}$. If the induced emf between the centre and edge of the blade is $0.02 \mathrm{~V}$, determine the rate of rotation of the blade.
Solution:
Length of the metal blade, $1=0.4 \mathrm{~m}$
Magnetic field, $\mathrm{B}=4 \times 10^{-3} \mathrm{~T}$
Induced emf, e $=0.02 \mathrm{~V}$
Rotational area of the blade, $\mathrm{A}=\pi \mathrm{r}^2=3.14 \times(0.4)^2=0.5024 \mathrm{~m}^2$
Induced emf in rotational of the coil, $\mathrm{e}=\mathrm{NBA} \omega \sin \theta$
$
\begin{aligned}
& \omega=\frac{e}{\mathrm{NBA} \sin \theta} \quad\left[\mathrm{N}=1, \theta=90^{\circ}, \sin 90^{\circ}=1\right] \\
& \omega=\frac{0.02}{1 \times 4 \times 10^{-3} \times 0.5024 \times \sin 90^{\circ}}=\frac{0.02}{2.0096 \times 10^{-3}} \\
& =9.95222 \times 10^{-3} \times 10^3 \\
& =9.95 \text { revolutions } / \text { second } \\
& \text { Rate of rotational of the blade }, \omega=9.95 \text { revolutions } / \text { second }
\end{aligned}
$

Question 8.
A bicycle wheel with metal spokes of $1 \mathrm{~m}$ long rotates in Earth's magnetic field. The plane of the wheel is perpendicular to the horizontal component of Earth's field of $4 \times 10^{-5} \mathrm{~T}$. If the emf induced across the spokes is $31.4 \mathrm{mV}$, calculate the rate of revolution of the wheel.
Solution:
Length of the metal spokes, $1=1 \mathrm{~m}$
Rotational area of the spokes, $\mathrm{A}=\pi^2=3.14 \mathrm{x}(1)^2=3.14 \mathrm{~m}^2$
Horizontal component of Earth's field, $B=4 \times 10^{-5} \mathrm{~T}$
Induced $\mathrm{emf}, \mathrm{e}=31.4 \mathrm{mV}$
The rate of revolution of wheel,
$
\begin{aligned}
& 1, \omega=\frac{e}{\mathrm{NBA} \sin \theta} \quad\left[\mathrm{N}=1, \theta=90^{\circ}, \sin 90^{\circ}=1\right] \\
& =\frac{31.4 \times 10^{-3}}{1 \times 4 \times 10^{-5} \times 3.14 \times \sin 90^{\circ}}=\frac{31.4 \times 10^{-3}}{12.56 \times 10^{-5}}=2.5 \times 10^2 \\
& \omega=250 \text { revolutions } / \text { second }
\end{aligned}
$
Question 9.
Determine the self-inductance of $4000 \mathrm{turn}$ air-core solenoid of length $2 \mathrm{~m}$ and diameter $0.04 \mathrm{~m}$.
Solution:
Length of the air core solenoid, $1=2 \mathrm{~m}$
Diameter, $\mathrm{d}=0.04 \mathrm{~m}$
Radius, $\mathrm{r}=\frac{d}{2}=0.02 \mathrm{~m}$
Area of the air core solenoid, $\mathrm{A}=\pi^2=3.14 \times(0.02)^2=1.256 \times 10^{-3} \mathrm{~m}^2$
Number of Turns, $\mathrm{N}=4000$ turns
Self inductance, $\mathrm{L}=\mu_0 \mathrm{n}^2 \mathrm{Al}$
$
\begin{aligned}
& =\mu_0 \frac{\mathrm{N}^2}{l^2} \times \mathrm{A} l \quad\left[n=\frac{\mathrm{N}}{l}, \mu_0=4 \pi \times 10^{-7} \mathrm{Hm}^{-1}\right] \\
& =\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l}=\frac{4 \pi \times 10^{-7} \times(4000)^2 \times 1.256 \times 10^{-3}}{2} \\
& =\frac{252405760 \times 10^{-10}}{2} \\
& =126202880 \times 10^{-10}=12.62 \times 10^{-3}=12.62 \mathrm{mH}
\end{aligned}
$

Question 10.
A coil of 200 turns carries a current of $4 \mathrm{~A}$. If the magnetic flux through the coil is $6 \times 10^{-5} \mathrm{~Wb}$, find the magnetic energy stored in the medium surrounding the coil.
Solution:
Number of turns of the coil, $N=200$
Current, $\mathrm{I}=4 \mathrm{~A}$
Magnetic flux through the coil, $\Phi=6 \times 10^{-5} \mathrm{~Wb}$
Energy stored in the coil, $\mathrm{U}=\frac{1}{2} \mathrm{LI}^2=\frac{1}{I 2}$
Self inductance of the coil, $\mathrm{L}=\frac{N \Phi}{I}$
$\mathrm{U}=\frac{1}{2} \frac{N \Phi}{I} \times \mathrm{I}^2=\frac{1}{2} \mathrm{~N} \Phi \mathrm{I}=\frac{1}{2} \times 200 \times 6 \times 10^{-5} \times 4$
$\mathrm{U}=2400 \times 10^{-5}=0.024 \mathrm{~J}$ (or) joules.
Question 11.
A $50 \mathrm{~cm}$ long solenoid has 400 turns per cm. The diameter of the solenoid is $0.04 \mathrm{~m}$. Find the magnetic flux of a turn when it carries a current of $1 \mathrm{~A}$.
Solution:
Length of the solenoid, $1=50 \mathrm{~cm}=50 \times 10^{-2} \mathrm{~m}$
Number of turns per $\mathrm{cm}, \mathrm{N}=400$
Number of turns in $50 \mathrm{~cm}, \mathrm{~N}=400 \times 50=20000$
Diameter of the solenoid, $\mathrm{d}=0.04 \mathrm{~m}$
Radius of the solenoid, $\mathrm{r}=\frac{d}{2}=0.02 \mathrm{~m}$
Area of the solenoid, $\mathrm{A}=\pi^2=3.14 \times(0.02)^2=1.256 \times 10^{-3} \mathrm{~m}^2$
Current passing through the solenoid, $\mathrm{I}=1 \mathrm{~A}$
Magnetic fluex,
$
\begin{aligned}
\phi & =\frac{\mu_0 \mathrm{~N}^2 \mathrm{AI}}{l}=\frac{4 \pi \times 10^{-7} \times(20000)^2 \times 1.256 \times 10^{-3} \times 1}{50 \times 10^{-2}} \\
& =\frac{4 \times 3.14 \times 4 \times 10^8 \times 10^{-10} \times 1.256}{50 \times 10^{-2}}=\frac{63.10144 \times 10^{-2}}{50 \times 10^{-2}} \\
\phi & =1.262 \mathrm{~Wb}
\end{aligned}
$

Question 12.
A coil of 200 turns carries a current of $0.4 \mathrm{~A}$. If the magnetic flux of $4 \mathrm{mWb}$ is linked with the coil, find the inductance of the coil.
Solution:
Number of turns, $\mathrm{N}=200$; Current, $\mathrm{I}=0.4 \mathrm{~A}$
Magnetic flux linked with coil, $\Phi=4 \mathrm{mWb}=4 \times 10^{-3} \mathrm{~Wb}$
Induction of the coil, $\mathrm{L}=\frac{N \Phi}{I}=\frac{200 \times 4 \times 10^{-3}}{0.4}=\frac{800 \times 10^{-3}}{0.4} 2 \mathrm{H}$
Question 13.
Two air core solenoids have the same length of $80 \mathrm{~cm}$ and same cross-sectional area $5 \mathrm{~cm}^2$. Find the mutual inductance between them if the number of turns in the first coil is 1200 turns and that in the second coil is 400 turns.
Solution:
Length of the solenoids, $1=80 \mathrm{~cm}=8 \times 10^{-2} \mathrm{~m}$
Cross sectional area of the solenoid, $\mathrm{A}=5 \mathrm{~cm}^2=5 \times 10^{-4} \mathrm{~m}^2$
Number of turns in the $\mathrm{I}^{\mathrm{st}}$ coil, $\mathrm{N}_1=1200$
Number of turns in the IInd coil, $\mathrm{N}_2=400$
Mutual inductance between the two coils,
$
\begin{aligned}
& \text { oils, } M=\frac{\mu_0 \mathrm{~N}_1 \mathrm{~A} l \mathrm{~N}_2}{l^2}=\frac{\mu_0 \mathrm{~N}_1 \mathrm{AN}_2}{l} \\
& =\frac{4 \pi \times 10^{-7} \times 1200 \times 400 \times 5 \times 10^{-4}}{80 \times 10^{-2}}=\frac{30144000 \times 10^{-11}}{80 \times 10^{-2}} \\
& =376800 \times 10^{-9}=0.38 \times 10^{-3}=0.38 \mathrm{mH}
\end{aligned}
$
Question 14.
A long solenoid having 400 turns per cm carries a current 2A. A 100 turn coil of cross sectional area 4 $\mathrm{cm}^2$ is placed co-axially inside the solenoid so that the coil is in the field produced by the solenoid. Find the emf induced in the coil if the current through the solenoid reverses its direction in 0.04 sec.

Solution:
Number of turns of long solenoid per $\mathrm{cm}=\frac{400}{10^{-2}} ; \mathrm{N}_2=400 \times 10^2$
Number of turns inside the solenoid, $\mathrm{N}_2=100$
Cross-sectional area of the coil, $\mathrm{A}=4 \mathrm{~cm}^2=4 \times 10^{-4} \mathrm{~m}^2$
Current through the solenoid, $I=2 \mathrm{~A}$; time, $\mathrm{t}=0.04 \mathrm{~s}$
Induced emf of the coil, e $=-\mathrm{M} \frac{d I}{d t}$
Mutual inductance of the coil,
$
\begin{aligned}
\mathrm{M} & =\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{~A}}{l}=\frac{4 \pi \times 10^{-7} \times 400 \times 10^2 \times 100 \times 4 \times 10^{-4}}{1} \\
& =2009600 \times 10^{-9}=200.96 \times 10^{-5}=2 \mathrm{mH}
\end{aligned}
$
Induced emf of the coil,
$
\begin{aligned}
e & =-\mathrm{M} \frac{d \mathrm{I}}{d t}=-2 \times 10^{-3} \times \frac{2}{0.04}=\frac{-4 \times 10^{-3}}{0.04} \\
& =-0.10=-0.1 \mathrm{~V}
\end{aligned}
$
The current through the solenoid reverse its direction if the induced emf, e $=-0.2 \mathrm{~V}$
Question 15.
A 200 turn coil of radius $2 \mathrm{~cm}$ is placed co-axially within a long solenoid of $3 \mathrm{~cm}$ radius. If the turn density of the solenoid is 90 turns per $\mathrm{cm}$, then calculate mutual inductance of the coil.
Solution:
Number of turns of the solenoid, $\mathrm{N}_2=200$
Radius of the solenoid, $\mathrm{r}=2 \mathrm{~cm}=2 \times 10^2 \mathrm{~m}$
Area of the solenoid, $A=\pi r^2=3.14 \times\left(2 \times 10^{-2}\right)^2=1.256 \times 10^{-3} \mathrm{~m}^2$
Turn density of long solenoid per $\mathrm{cm}, \mathrm{N}_1=90 \times 10^2$
Mutual inductance of the coil,
$
\begin{aligned}
& \mathrm{M}=\frac{\mu_0 \mathrm{~N}_1 \mathrm{~N}_2 \mathrm{~A}}{l}=\frac{4 \pi \times 10^{-7} \times 90 \times 10^2 \times 200 \times 1.256 \times 10^{-3}}{1} \\
& =283956.48 \times 10^{-8} \Rightarrow \mathrm{M}=2.84 \mathrm{mH}
\end{aligned}
$
Question 16.
The solenoids $S_1$ and $S_2$ are wound on an iron-core of relative permeability 900 . The area of their cross-section and their length are the same and are $4 \mathrm{~cm}^2$ and $0.04 \mathrm{~m}$, respectively. If the number of turns in $\mathrm{S}_1$ is 200 and that in $\mathrm{S}_2$ is 800, calculate the mutual inductance between the coils. The current in solenoid 1 is increased from $2 \mathrm{~A}$ to $8 \mathrm{~A}$ in 0.04 second. Calculate the induced emf in solenoid 2.
Solution:
Relative permeability of iron core, $\mu_{\mathrm{r}}=900$
Number of turns of solenoid $\mathrm{S}_1, \mathrm{~N}_1=200$

Number of turns of solenoid $S_2, N_2=' 800$
Area of cross section, $\mathrm{A}=4 \mathrm{~cm}^2=4 \times 10^{-4} \mathrm{~m}^2$
Length of the solenoid $\mathrm{S}_1, 1_1=0.04 \mathrm{~m}$
current, $I=I_2-I_1=8-2=6 \mathrm{~A}$
time taken, $t=0.04$ second
$\mathrm{emf}$ induced in solenoid $\mathrm{S}_2 \mathrm{e}=-\mathrm{M} \frac{d I}{d t}$
Mutual inductance between the two coils, $\mathrm{M}=\frac{\mu_0 \mu_r N_1 N_2 A}{l}$
$=\frac{4 \pi \times 10^{-7} \times 900 \times 200 \times 800 \times 4 \times 10^{-4}}{0.04}$
$=\frac{4 \times 3.14 \times 10^{-7} \times 576 \times 10^6 \times 10^{-4}}{0.04}=\frac{7234.56 \times 10^{-5}}{0.04}$
$\mathrm{M}=180864 \times 10^{-5}=1.81 \mathrm{H}$
Emf induced in solenoid $\mathrm{S}_2, \mathrm{e}=-\mathrm{M} \frac{d I}{d t}=-1.81 \times \frac{6}{0.04}$
Magnitude of emf, e $=271.5 \mathrm{~V}$
Question 17.
A step-down transformer connected to main supply of $220 \mathrm{~V}$ is made to operate $11 \mathrm{~V}, 88 \mathrm{~W}$ lamp.
Calculate (i) Transformation ratio and (ii) Current in the primary.
Solution:
Voltage in primary coil, $\mathrm{V}_{\mathrm{p}}=220 \mathrm{~V}$
Voltage in secondary coil, $\mathrm{V}_{\mathrm{s}}=11 \mathrm{~V}$
Output power $=88 \mathrm{~W}$
(i) To find transformation ratio, $\mathrm{k}=\frac{V_s}{V_p}=\frac{11}{220}=\frac{1}{20}$
(ii) Current in primary, $\mathrm{I}_{\mathrm{p}}=\frac{V_s}{V_p} \times \mathrm{I}_{\mathrm{s}}$
So, $\mathrm{I}_{\mathrm{s}}=$ ?
Outputpower $=\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{S}}$
$\Rightarrow 88=11 \times \mathrm{I}_{\mathrm{s}}$
$\mathrm{I}_{\mathrm{s}}=\frac{88}{11}=8 \mathrm{~A}$
Therefore, $\mathrm{I}_{\mathrm{p}}=\frac{V_s}{V_p} \times \mathrm{I}_{\mathrm{s}}=\frac{11}{220} \times 8=0.4 \mathrm{~A}$

Question 18.
A $200 \mathrm{~V} / 120 \mathrm{~V}$ step-down transformer of $90 \%$ efficiency is connected to an induction stove of resistance $40 \Omega$. Find the current drawn by the primary of the transformer.
Solution:
Primary voltage, $\mathrm{V}_{\mathrm{p}}=200 \mathrm{~V}$
Secondary voltage, $\mathrm{V}_{\mathrm{S}}=120 \mathrm{~V}$
Efficiency, $\eta=90 \%$
Secondary resistance, $\mathrm{R}_{\mathrm{s}}=40 \Omega$
Current drawn by the primary of the transforme, $\mathrm{I}_{\mathrm{p}}=\frac{V_s}{R_s} \times \mathrm{I}_{\mathrm{s}}$
Output power $=\mathrm{V}_{\mathrm{S}} \mathrm{I}_{\mathrm{s}}$
From Ohm's law, $\mathrm{I}_s=\frac{\mathrm{V}_s}{\mathrm{R}_s}=\frac{\mathrm{V}_s^2}{\mathrm{R}_s}=\frac{(120)^2}{40}=360 \mathrm{~W}$
$
\therefore \mathrm{I}_s=\frac{\text { Output power }}{\mathrm{V}_s}=\frac{360}{120}=3 \mathrm{~A}
$
Efficiency, $\eta=\frac{\text { Output power }}{\text { Input power }}=\frac{\mathrm{V}_s \mathrm{I}_s}{\mathrm{~V}_p \mathrm{I}_p}$
$
\mathrm{I}_p=\frac{\mathrm{V}_s \mathrm{I}_s}{\mathrm{~V}_p \eta}=\frac{120 \times 3 \times 100}{200 \times 90}=2 \mathrm{~A}
$

Question 19.
The 300 turn primary of a transformer has resistance $0.82 \Omega$ and the resistance of its secondary of 1200 turns is $6.2 \Omega$. Find the voltage across the primary if the power output from the secondary at $1600 \mathrm{~V}$ is $32 \mathrm{~kW}$. Calculate the power losses in both coils when the transformer efficiency is $80 \%$.
Solution:
Efficiency, $\eta=80 \%=\frac{80}{100}$
Number of turns in primary, $\mathrm{N}_{\mathrm{p}}=300$
Number of turns in secondary, $\mathrm{N}_{\mathrm{s}}=1200$
Resistance in primary, $\mathrm{R}_{\mathrm{p}}=0.82 \Omega$
Resistance in secondary, $\mathrm{R}_{\mathrm{s}}=6.2 \Omega$
Secondary voltage, $\mathrm{V}_{\mathrm{S}}=1600 \mathrm{~V}$
Output power $=32 \mathrm{~kW}$
Output power $=V_{\mathrm{S}} \mathrm{I}_{\mathrm{S}}$
$
\mathrm{I}_s=\frac{\text { Output power }}{\mathrm{V}_s}=\frac{32 \times 10^3}{1600}=20 \mathrm{~A}
$
Efficiency, $\eta=\frac{\text { Output power }}{\text { Input power }}$
$
\frac{80}{100}=\frac{32 \times 10^3}{\text { Input power }}
$
Input power $=40 \mathrm{~kW}$
$
\begin{aligned}
\frac{\mathrm{N}_s}{\mathrm{~N}_p} & =\frac{\mathrm{V}_s}{\mathrm{~V}_p} \quad \Rightarrow \mathrm{V}_p=\frac{\mathrm{V}_s \mathrm{~N}_p}{\mathrm{~N}_s}=\frac{1600 \times 300}{1200} \\
\mathrm{~V}_p & =400 \mathrm{~V} \\
\text { Input power } & =\mathrm{V}_p \mathrm{I}_p \\
\mathrm{I}_p & =\frac{\text { Input power }}{\mathrm{V}_p}=\frac{40000}{400}=100 \mathrm{~A}, \quad\left[\mathrm{I}_p=100 \mathrm{~A}\right]
\end{aligned}
$

Power loss in primary $=I_p^2 R_p=(100)^2 \times 0.82=8200=8.2 \mathrm{~kW}$
Power loss in secondary $=I_s^2 R_s=(20)^2 \times 6.2=2480=2.48 \mathrm{~kW}$
Question 20.
Calculate the instantaneous value at $60^{\circ}$, average value and RMS value of an alternating current whose peak value is $20 \mathrm{~A}$.
Solution:
Peak value of current, $\mathrm{I}_{\mathrm{m}}=20 \mathrm{~A}$
Angle, $\theta=60^{\circ}[\theta=\omega \mathrm{t}]$
(i) Instantaneous value of current,
$
\begin{aligned}
& \mathrm{i}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}=\mathrm{I}_{\mathrm{m}} \sin \theta \\
& =20 \sin 60^{\circ}=20 \times \frac{\sqrt{ } 3}{2}=10 \sqrt{ } 3=10 \times 1.732 \\
& \mathrm{i}=17.32 \mathrm{~A}
\end{aligned}
$
(ii) Average value of current,
$
\begin{aligned}
& \mathrm{I}_{\mathrm{av}}=\frac{2 I_m}{\pi}=\frac{2 \times 20}{3.14} \\
& \mathrm{I}_{\mathrm{av}}=12.74 \mathrm{~A}
\end{aligned}
$
(iii) RMS value of current,
$\mathrm{I}_{\mathrm{RMS}}=0.707 \mathrm{I}_{\mathrm{m}}$
or $\frac{\mathrm{I}_{\mathrm{m}}}{\sqrt{2}}=0.707 \times 20$
$\mathrm{I}_{\mathrm{RMS}}=14.14 \mathrm{~A}$