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

Updated On May 15, 2024

By SaraNextGen

Additional Questions solved
I Choose The Correct Answer
Question 1.
A coil of area of cross section $0.5 \mathrm{~m}^2$ with 10 turns is in a plane which is perpendicular to an uniform magnetic field of $0.2 \mathrm{~Wb} / \mathrm{m}^2$. The flux through the coil is -
(a) $100 \mathrm{~Wb}$
(b) $10 \mathrm{~Wb}$
(c) $1 \mathrm{~Wb}$
(d) zero
Answer:
(c) $1 \mathrm{~Wb}$
Hint:
$
\begin{aligned}
& \Phi=\mathrm{NBA} \cos \theta \\
& =10 \times 0.2 \times 0.5 \times \cos 0^{\circ}=1 \mathrm{~Wb}
\end{aligned}
$
Question 2.
A rectangular coil of 100 turns and size $0.1 \mathrm{~m} \mathrm{x} 0.05 \mathrm{~m}$ is placed perpendicular to a magnetic field of $0.1 \mathrm{~T}$. If the field drops to $0.05 \mathrm{~T}$ in $0.05 \mathrm{~s}$, the magnitude of the emf induced in the coil is-
(a) $0.5 \mathrm{~V}$
(b) $0.75 \mathrm{~V}$
(c) $1.0 \mathrm{~V}$
(d) $1.5 \mathrm{~V}$
Answer:
(a) $0.5 \mathrm{~V}$
Hint:
$
\begin{aligned}
& \varepsilon=-\frac{\mathrm{NA}\left(\mathrm{B}_f-\mathrm{B}_i\right)}{t}=-\frac{100 \times 0.1 \times 0.05 \times(0.05-0.1)}{0.05} \\
& \varepsilon=0.5 \mathrm{~V}
\end{aligned}
$
Question 3
A wire of length $1 \mathrm{~m}$ moves with a speed of $10 \mathrm{~ms}^{-1}$ perpendicular to a magnetic field. If the emf induced in the wire is $1 \mathrm{~V}$, the magnitude of the field is-
(a) $0.01 \mathrm{~T}$
(b) $0.1 \mathrm{~T}$
(c) $0.2 \mathrm{~T}$
(d) $0.02 \mathrm{~T}$
Answer:
(b) $0.1 \mathrm{~T}$
Hint:

$
\begin{aligned}
& \varepsilon=\mathrm{Blv} \\
& \Rightarrow \mathrm{B}=\frac{\varepsilon}{l v}=\frac{1}{1 \times 10}=0.02 \mathrm{~T}
\end{aligned}
$
Question 4.
A coil of area $10 \mathrm{~cm}^2, 10 \mathrm{~ms}^{-1}$ turns and resistance $20 \Omega$ is placed in a magnetic field directed
perpendicular to the plane of the coil and changing at the rate of $10^8$ gauss $/$ second. The induced current in the coil will be-
(a) $5 \mathrm{~A}$
(b) $0.5 \mathrm{~A}$
(c) $0.05 \mathrm{~A}$
(d) $50 \mathrm{~A}$
Answer:
(a) $5 \mathrm{~A}$
Hint:
$
\begin{array}{rlr}
\varepsilon & =\frac{d}{d t}(\mathrm{NAB})=\mathrm{NA} \frac{d \mathrm{~B}}{d t} ; & \frac{d \mathrm{~B}}{d t}=10^8 \text { gauss/second } \\
\varepsilon & =10 \times 10 \times 10^{-4} \times 10^4 ; & \frac{d \mathrm{~B}}{d t}=10^4 \mathrm{~T} / \mathrm{second} \\
\varepsilon & =100 \mathrm{~V} \\
\mathrm{I} & =\frac{\varepsilon}{\mathrm{R}}=\frac{100}{20}=5 \mathrm{~A} &
\end{array}
$
Question 5.
A coil of cross sectional area $400 \mathrm{~cm}^2$ having 30 turns is making $1800 \mathrm{rev} / \mathrm{min}$ in a magnetic field of IT. The peak value of the induced emf is-
(a) $113 \mathrm{~V}$
(b) $226 \mathrm{~V}$
(c) $339 \mathrm{~V}$
(d) $452 \mathrm{~V}$
Answer:
(b) $226 \mathrm{~V}$
Hint:

$
\begin{aligned}
& \varepsilon_{\mathrm{m}}=\mathrm{NBA} \omega=30 \times 1 \times 400 \times 10^{-4} \times 30 \times 2 \pi \\
& =226 \mathrm{~V}
\end{aligned}
$
Question 6.
Eddy currents are produced in a material when it is-
(a) heated
(b) placed in a time varying magnetic field
(c) placed in an electric field
(d) placed a uniform magnetic field
Answer:
(b) placed in a time varying magnetic field
Question 7.
An emf of $5 \mathrm{~V}$ is induced in an inductance when the current in it changes at a steady rate from $3 \mathrm{~A}$ to $2 \mathrm{~A}$ in 1 millisecond. The value of inductance is-
(a) $5 \mathrm{mH}$
(b) $5 \mathrm{H}$
(c) $5000 \mathrm{H}$
(d) zero
Answer:
(a) $5 \mathrm{mH}$

Question 8.
Faraday's law of electromagnetic induction is related to the-
(a) Law of conservation of charge
(b) Law of conservation of energy
(c) Third law of motion
(d) Law of conservation of angular momentum
Answer:
(b) Law of conservation of energy
Question 9.
The inductance of a coil is proportional to-
(a) its length
(b) the number of turns
(c) the resistance of the coil
(d) square of the number of turns
Answer:
(d) square of the number of turns
Question 10.
When a direct current ' $i$ ' is passed through an inductance $\mathrm{L}$, the energy stored is-
(a) Zero
(b) $\mathrm{Li}$
(c) $\frac{1}{2} \mathrm{Li}^2$
(d) $\frac{L^2}{2 i}$
Answer:
(c) $\frac{1}{2} \mathrm{Li}^2$
Question 11.
A coil of area $80 \mathrm{~cm}^2$ and 50 turns is rotating with 2000 revolutions per minute about an axis perpendicular to a magnetic field of $0.05 \mathrm{~T}$. The maximum value of the emf developed in it is-
(a) $2000 \pi \mathrm{V}$
(b) $\frac{10 \pi}{3} \mathrm{~V}$
(c) $\frac{4 \pi}{3} \mathrm{~V}$
(d) $\frac{2}{3} \mathrm{~V}$
Answer:
(c) $\frac{4 \pi}{3} \mathrm{~V}$
Hint:
$
\varepsilon=\mathrm{NBA} \omega=50 \times 0.05 \times 80 \times 10^{-4} \times \frac{2 \pi \times 2000}{60}=\frac{4 \pi}{3} \mathrm{~V}
$

Question 12 .
The direction of induced current during electro magnetic induction is given by-
(a) Faraday's law
(b) Lenz's law
(c) Maxwell's law
(d) Ampere's law
Answer:
(b) Lenz's law
Question 13.
AC power is transmitted from a power house at a high voltage as-
(a) the rate of transmission is faster at high voltages
(b) it is more economical due to less power loss
(c) power cannot be transmitted at low voltages
(d) a precaution against theft of transmission lines
Answer:
(b) it is more economical due to less power loss
Question 14.
In a step-down transformer the input voltage is $22 \mathrm{kV}$ and the output voltage is $550 \mathrm{~V}$. The ratio of the number of turns in the secondary to that in the primary is-
(a) $1: 20$
(b) $20: 1$
(c) $1: 40$
(d) $40: 1$
Answer:
(c) $1: 40$
Hint:
$
\frac{N_s}{N_p}=\frac{V_s}{V_p}=\frac{550}{22000}=\frac{1}{40}
$
Question 15 .
The self-inductance of a coil is $5 \mathrm{H}$. A current of $1 \mathrm{~A}$ changes to $2 \mathrm{~A}$ within $5 \mathrm{~s}$ through the coil. The value of induced emf will be-
(a) $10 \mathrm{~V}$
(b) $0.1 \mathrm{~V}$
(c) $1.0 \mathrm{~V}$
(d) $100 \mathrm{~V}$
Answer:

(c) $1.0 \mathrm{~V}$
Hint:
$
|\varepsilon|=\left|\mathrm{L} \frac{d i}{d t}\right|=5 \times\left(\frac{2-1}{5}\right)=1.0 \mathrm{~V}
$
Question 16.
The low-loss transformer has $230 \mathrm{~V}$ applied to the primary and gives $4.6 \mathrm{~V}$ in the secondary. The secondary is connected to a load which draws 5 amperes of current. The current (in amperes) in the primary is-
(a) $0.1 \mathrm{~A}$
(b) $1.0 \mathrm{~A}$
(c) $10 \mathrm{~A}$
(d) $250 \mathrm{~A}$
Answer:
(a) $0.1 \mathrm{~A}$
Hint:
$
\mathrm{I}_{\mathrm{p}}=\frac{V_s I_s}{V_p}=\frac{4.6 \times 5}{230}=0.1 \mathrm{~A}
$

Question 17.
A coil is wound on a frame of rectangular cross-section. If all the linear dimensions of the frame are increased by a factor 2 and the number of turns per unit length of the coil remains the same. Selfinductance of the coil increases by a factor of-
(a) 4
(b) 8
(c) 12
(d) 16
Answer:
(b) 8
Hint:
If all the linear dimensions are doubled, the cross-sectional are a becomes eight times. Therefore, the flux produced by a given current will become eight times. Hence, the selfinductance increases by a factor of 8 .
Question 18.
If $\mathrm{N}$ is the number of turns in a coil, the value of self-inductance varies as-
(a) $\mathrm{N}^{\circ}$
(b) $\mathrm{N}$
(c) $\mathrm{N}^2$
(d) $\mathrm{N}^{-2}$
Answer:
(c) $\mathrm{N}^2$
Hint:
According to self inductance of long solenoid
$
\begin{aligned}
& \mathrm{L}=\frac{\mu_0 \mathrm{~N}^2 \mathrm{~A}}{l} \\
& \Rightarrow \mathrm{L} \propto \mathrm{N}^2
\end{aligned}
$
Question 19.
A magnetic field $2 \times 10^{-2} \mathrm{~T}$ acts at right angles to a coil of area $100 \mathrm{~cm}^2$ with 50 turns. The average emf induced in the coil will be $0.1 \mathrm{~V}$ if it is removed from the field in time.
(a) $0.01 \mathrm{~s}$
(b) $0.1 \mathrm{~s}$
(c) $1 \mathrm{~s}$
(d) $10 \mathrm{~s}$
Answer:
(b) $0.1 \mathrm{~s}$

Hint:
$
\begin{aligned}
\varepsilon & =\mathrm{NA} \frac{\Delta \mathrm{B}}{\Delta t} \\
\Delta t & =\frac{\mathrm{NA} \Delta \mathrm{B}}{\varepsilon}=\frac{50 \times 100 \times 10^{-4} \times 2 \times 10^{-2}}{0.1}=0.1 \mathrm{~s}
\end{aligned}
$
Question 20.
Number of turns in a coil is increased from 10 to 100 . Its inductance becomes-
(a) 10 times
(b) 100 times
(c) $1 / 10$ times
(d) 25 times
Answer:
(a) 10 times
Question 21.
The north pole of a magnet is brought near a metallic ring as shown in the figure. The direction of induced current in the ring, as seen by the magnet is-

(a) anti-clockwise
(b) first anti-clockwise and then clockwise
(c) clockwise
(d) first clockwise and then anti-clockwise
Answer:
(a) anti-clockwise
Question 22 .
Quantity that remains unchanged in a transformer is-
(a) voltage
(b) current
(c) frequency
(d) none of these
Answer:
(c) frequency
Question 23.
The core of a transformer is laminated to reduce.
(a) Copper loss
(b) Magnetic loss
(c) Eddy current loss
(d) Hysteresis loss
Answer:
(c) Eddy current loss
Question 24.
Which of the following has the dimension of time?
(a) $\mathrm{LC}$
(b) $\frac{R}{L}$
(c) $\frac{L}{R}$
(d) $\frac{C}{L}$
Answer:
(c) $\frac{L}{R}$
Question 25.
A coil has a self-inductance of $0.04 \mathrm{H}$. The energy required to establish a steady-state current of $5 \mathrm{~A}$ in it is-

(a) $0.5 \mathrm{~J}$
(b) $1.0 \mathrm{~J}$
(c) $0.8 \mathrm{~J}$
(d) $0.2 \mathrm{~J}$
Answer:
(a) $0.5 \mathrm{~J}$
Question 26.
Alternating current can be measured by
(a) moving coil galvanometer
(b) hot wire ammeter
(c) tangent galvanometer
(d) none of the above
Answer:
(b) hot wire ammeter
Question 27.
In an LCR circuit, the energy is dissipated in-
(a) R only
(b) R and L only
(c) R and C only
(d) R, L and C
Answer:
(a) R only
Question 28.
A $40 \Omega$ electric heater is connected to $200 \mathrm{~V}, 50 \mathrm{~Hz}$ main supply. The peak value of the electric current flowing in the circuit is approximately-
(a) $2.5 \mathrm{~A}$
(b) $5 \mathrm{~A}$
(c) $7 \mathrm{~A}$
(d) $10 \mathrm{~A}$
Answer:
(c) $7 \mathrm{~A}$
Hint:
$
\mathrm{I}_0=\frac{V_0}{R}=\frac{200 \sqrt{2}}{40} 5 \sqrt{2} \approx 7 \mathrm{~A}
$

Question 29.
The rms value of an alternating current, which when passed through a resistor produces heat three times of that produced by a direct current of $2 \mathrm{~A}$ in the same resistor, is-
(a) $6 \mathrm{~A}$
(b) $3 \mathrm{~A}$
(c) $2 \mathrm{~A}$
(d) $2 \sqrt{3} \mathrm{~A}$
Answer:
(d) $2 \sqrt{3} \mathrm{~A}$
Hint:
$
I_{r m s}^2 \mathrm{R}=3\left(2^2 \mathrm{R}\right) \text { (or) } \mathrm{I}_{\mathrm{rms}}=2 \sqrt{3} \mathrm{~A}
$
Question 30 .
An inductance, a capacitance and a resistance are connected in series across a source of alternating voltages. At resonance, the applied voltage and the current flowing through the circuit will have a phase difference of-
(a) $\frac{\pi}{4}$
(b) zero
(c) $\pi$
(d) $\frac{\pi}{2}$
Answer:
(b) zero
Question 31.
In an $\mathrm{AC}$ circuit, the rms value of the current $\mathrm{I}_{\text {rms }}$, is related to the peak current $\mathrm{I}_0$ as-
(a) $\mathrm{I}_{\mathrm{rms}}=\frac{I_0}{\pi}$
(b) $I_{\mathrm{rms}}=\frac{I_0}{\sqrt{ } 2}$
(c) $\mathrm{I}_{\mathrm{rms}}=\sqrt{2} \mathrm{I}_0$
(d) $\mathrm{I}_{\mathrm{rms}}=\pi \mathrm{I}_0$
Answer:
(b) $\mathrm{I}_{\mathrm{rms}}=\frac{I_0}{\sqrt{ } 2}$
Question 32.
The impedance of a circuit consists of $3 \Omega$ resistance and $4 \Omega$ resistance. The power factor of the circuit is
(a) 0.4
(b) 0.6
(c) 0.8
(d) 1.0
Answer:

(b) 0.6
Hint:
$
\tan \Phi=\frac{4}{3}
$
Power factor $=\cos \Phi=\frac{3}{5}=0.06$
Question 33.
The reactance of a capacitance at $50 \mathrm{~Hz}$ is $5 \Omega$. Its reactance at $100 \mathrm{~Hz}$ will be-
(a) $5 \Omega$
(b) $10 \Omega$
(c) $20 \Omega$
(d) $2.5 \Omega$
Answer:
(d) $2.5 \Omega$.
Question 34 .
In a LCR AC circuit off resonance, the current-
(a) is always in phase with the voltage
(b) always lags behind the voltage
(c) always leads the voltage
(d) may lead or lag behind the voltage
Answer:
(d) may lead or lag behind the voltage
Question 35.
The average power dissipation in a pure inductance $\mathrm{L}$, through which a current $\mathrm{I}_0 \sin \omega t$ is flowing is-
(a) $\frac{1}{2} \mathrm{~L} I_0^2$
(b) $\mathrm{L} I_0^2$
(c) $2 \mathrm{~L} I_0^2$
(d) zero
Answer:
(d) zero

Question 36 .
The power in an $\mathrm{AC}$ circuit is given by $P=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \Phi$. The value of the power factor $\cos \Phi$ in series LCR circuit at resonance is-
(a) zero
(b) 1
(c) $\frac{1}{2}$
(d) $\frac{1}{\sqrt{ } 2}$
Answer:
(b) 1
Hint:
$
\begin{aligned}
& \cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{\mathrm{R}}{\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right)}} \quad \text { at resonance } \mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}} \\
& \cos \phi=\frac{\mathrm{R}}{\mathrm{R}}=1
\end{aligned}
$

Question 37.
In an $\mathrm{AC}$ circuit with voltage $\mathrm{V}$ and current $\mathrm{I}$, the power dissipated is-
(a) VI
(b) $\frac{1}{2} \mathrm{VI}$
(c) $\frac{1}{\sqrt{ } 2} \mathrm{VI}$
(d) depends on the phase difference between $\mathrm{I}$ and $\mathrm{V}$
Answer:
(d) depends on the phase difference between $\mathrm{I}$ and $\mathrm{V}$
Question 38.
In an $\mathrm{AC}$ circuit containing only capacitance, the current-
(a) leads the voltage by $180^{\circ}$
(b) remains in phase with the voltage
(c) leads the voltage by $90^{\circ}$
(d) lags the voltage by $90^{\circ}$
Answer:
(c) leads the voltage by $90^{\circ}$
Question 39.
In a series LCR circuit $R=10 \Omega$ and the impedance $Z=20 \Omega$. Then the phase difference between the current and the voltage is-
(a) $60^{\circ}$
(b) $30^{\circ}$
(c) $45^{\circ}$
(d) $90^{\circ}$
Answer:
(c) $60^{\circ}$
Hint:
$
\begin{aligned}
& \cos \Phi=\frac{R}{Z}=\frac{10}{20}=\frac{1}{2} \\
& \Rightarrow \Phi=60^{\circ}
\end{aligned}
$

Question 40.
What is the value of inductance $\mathrm{L}$ for which the current is maximum in a series $L C R$ circuit with $C=$ $10 \mu \mathrm{F}$ and $\omega=1000 \mathrm{~s}^{-1}$ ?
(a) $1 \mathrm{mH}$
(b) $10 \mathrm{mH}$
(c) $100 \mathrm{mH}$
(d) Cannot be calculated unless $\mathrm{R}$ is known
Answer:
(c) $100 \mathrm{mH}$
Hint:
$
\mathrm{L}=\frac{1}{\omega^2 C}=\frac{1}{(1000)^2 \times 10 \times 10^{-6}}=0.1 \mathrm{H}=100 \mathrm{mH}
$
II Fill in the Blanks
Question 1.
Electromagnetic induction is used in ..............
Answer:
transformer and $\mathrm{AC}$ generator .................
Question 2.
Lenz's Law is in accordance with the law of ..................
Answer:
conservation of energy.
Question 3.
The self-inductance of a straight conductor is ....................
Answer:
zero
Question 4.
Transformer works on ....................
Answer:
AC only
Question 5.
The power loss is less in transmission lines when ..................
Answer:
voltage is more but current is less

Question 6.
The law that gives the direction of the induced current produced in a circuit is ................
Answer:
Lenz's law
Question 7.
Fleming's right hand rule is otherwise called ..............
Answer:
generator rule
Question 8.
Unit of self-inductance is ............
Answer:
Henry
Question 9.
The mutual induction is very large, if the two coils are wound on ..........
Answer:
soft iron core
Question 10.
When the coil is in vertical position, the angle between the normal to the plane of the coil and magnetic field is .........
Answer:
zero

Question 11.
The emf induced by changing the orientation of the coil is ...................... in nature.
Answer:
sinusoidal
Question 12.
In a three phase $\mathrm{AC}$ generator, the three coils are inclined at an angle of ..............

Answer:

$120^{\circ}$
Question 13.
The emf induced in each of the coils differ in phase by ..............
Answer:
$120^{\circ}$
Question 14.
A device which converts high alternating voltage into low alternating voltage and vice versa is ...........
Answer:
transformer
Question 15.
For an ideal transformer efficiency $\eta$ is ..............
Answer:
1
Question 16.
The alternating emf induced in the coil varies ..................
Answer:
periodically in both magnitude and direction
Question 17.
For direct current, inductive reactance is .....................
Answer:
zero
Question 18.
In an inductive circuit the average power of sinusoidal quantity of double the frequency over a complete cycle is ...........

Answer:
zero

Question 19.
For direct current, the resistance offered by a capacitor is ................
Answer:
infinity
Question 20.
In a capacitive circuit, power over a complete cycle is .............
Answer:
zero
Question 21.
Q-factor measures the in resonant circuit ................
Answer:
selectivity
Question 22.
Voltage drop across inductor and capacitor differ in phase by ............
Answer:
$180^{\circ}$
Question 23.
Angular resonant frequency $(\mathrm{co})$ is ............
Answer:
$\frac{1}{\sqrt{L C}}$
Question 24.
A circuit will have flat resonance if its $\mathrm{Q}$-value is...............
Answer:
low
Question 25.
The average power consumed by the choke coil over a complete cycle is ...........
Answer:
zero
III Match the following
Question 1.

Answer:
(i) $\rightarrow$ (c)
(ii) $\rightarrow$ (a)
(iii) $\rightarrow$ (d)
(iv) $\rightarrow$ (b)
Question 2.

Answer:
(i) $\rightarrow$ (c)
(ii) $\rightarrow$ (a)
(iii) $\rightarrow$ (d)
(iv) $\rightarrow$ (b)
Question 3.

Answer:
(i) $\rightarrow$ (c)
(ii) $\rightarrow$ (d)
(iii) $\rightarrow$ (b)
(iv) $\rightarrow$ (a)
Question 4.
Type of impedance Phase between voltage and current

Answer:
(i) $\rightarrow$ (c)
(ii) $\rightarrow$ (d)
(iii) $\rightarrow$ (a)
(iv) $\rightarrow$ (b)
Question 5.
Energy in two oscillatory systems: (LC oscillator and spring mass system)

Answer:
(i) $\rightarrow$ (c)
(ii) $\rightarrow(\mathrm{d})$
(iii) $\rightarrow($ a $)$
(iv) $\rightarrow$ (b)
IV Assertion and reason
(a) If both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
(c) If assertion is true but reason is false.
(d) If the assertion and reason both are false.
(e) If assertion is false but reason is true.
Question 1.
Assertion: Eddy currents is produced in any metallic conductor when flux is changed around it. Reason: Electric potential determines the flow of charge.
Answer:
(b) If both assertion and reason are true but reason is not the correct explanation of the assertion.
Solution:
When a metallic conductor is moved in a magnetic field, magnetic flux is "varied. It disturbs the free electrons of the metal and set up an induced emf in it. As there are no free ends of the metal i.e., it will be closed in itself so there will be induced current.
Question 2.
Assertion: Faraday's laws are consequences of conservation of energy.
Reason: In a purely resistive AC circuit, the current lags behind the emf in phase.
Answer:
(c) If assertion is true but reason is false.
Solution:
According to Faraday's law, the conversion of mechanical energy into electrical energy is in accordance with the law of conservation of energy. It is also clearly known that in pure resistance, the $\mathrm{emf}$ is in phase with the current.
Question 3.
Assertion: Inductance coil are made of copper.

Reason: Induced current is more in wire having less resistance.
Answer:
(a) If both assertion and reason are true and the reason in the correct explanation of the assertion.
Solution:
Inductance coils made of copper will have very small ohmic resistance.
Question 4.
Assertion: An aircraft flies along the meridian, the potential at the ends of its wings will be the same. Reason: Whenever there is a change in the magnetic flux, and emf is induced.
Answer:
(e) If assertion is false but reason is true.
Solution:
As the aircraft flies magnetic flux change through its wings due to the vertical component of the Earth's magnetic field. Due to this, induced emf is produced across the wings of the aircraft. Therefore, the wings of the aircraft will not be at the same potential.
Question 5.
Assertion: In series LCR circuit resonance can take place.
Reason: Resonance takes place if inductance and capacitive reactances are equal and opposite.
Answer:
(a) If both assertion and reason are hue and the reason is the correct explanation of the assertion.
Solution:
At resonant frequency $\mathrm{X}_{\mathrm{L}}=\mathrm{X}_{\mathrm{C}}$
So, Impedance, $\mathrm{Z}=\mathrm{R}$ (minimum)
Therefore, the current in the circuit is maximum.
ShortAnswer Questions
Question 1.
Define magnetic flux $\left(\Phi_{\mathrm{B}}\right)$.

Answer:
The magnetic flux through an area $\mathrm{A}$ in a magnetic field is defined as the number of magnetic field lines passing through that area normally and is given by the equation,
$
\phi_{\mathrm{B}}=\int_{\mathrm{A}} \overrightarrow{\mathrm{B}} \cdot d \overrightarrow{\mathrm{A}}=\mathrm{BA} \cos \theta
$

Question 2.
Write down the drawbacks of Eddy currents.
Answer:
When eddy currents flow in the conductor, a large amount of energy is dissipated in the form of heat. The energy loss due to the flow of eddy current is inevitable but it can be reduced to a greater extent with suitable measures. The design of transformer core and electric motor armature is crucial in order to minimise the eddy current loss.
To reduce these losses, the core of the transformer is made up of thin laminas insulated from one another while for electric motor the winding is made up of a group of wires insulated from one another. The insulation used does not allow huge eddy currents to flow and hence losses are minimized.
Question 3.
Define the unit of self-inductance.
Answer:
The unit of self-inductance is henry. One henry is defined as the self-inductance of a coil in which a change in current of one ampere per second produces an opposing emf of one volt.
Question 4.
Define mutual inductance in terms of flux and current.
Answer:
The mutual inductance $\mathrm{M}_{21}$ is defined as the flux linkage of the coil 2 when 1A current flows through coil 1.
$
\mathrm{M}_{21}=\frac{\mathrm{N}_2 \phi_{21}}{i_1}
$
Question 5.
Define mutual inductance in terms of emf and current.
Answer:
Mutual inductance $\mathrm{M}_{21}$ is also defined as the opposing emf induced is the coil 2 when the rate of change of current through the coil 1 is $1 \mathrm{As}^{-1}$.
$
\mathbf{M}_{12}=\frac{-\varepsilon_1}{d i_2 / d t}
$

Question 6 .
List out the advantages of three phase alternator.
Answer:
Three-phase system has many advantages over single-phase system, which is as follows:
(i) For a given dimension of the generator, three-phase machine produces higher power output than a single-phase machine.
(ii) For the same capacity, three-phase alternator is smaller in size when compared to single phase alternator.
(iii) Three-phase transmission system is cheaper. A relatively thinner wire is sufficient for transmission of three-phase power.
Question 7.
Mentions the differences betw een a step up and step down transformer.
Answer:

Question 8.
Define efficiency of transformer.
Answer:
The efficiency $p$ of a transformer is defined as the ratio of the useful output power to the input power. Thus
$
\eta=\frac{\text { Output power }}{\text { Input power }} \times 100 \%
$
Transformers are highly efficient devices having their efficiency in the range of $96-99 \%$. Various energy losses in a transformer will not allow them to be $100 \%$ efficient.
Question 9.
What is meant by sinusoidal alternating voltage?
Answer:
If the waveform of alternating voltage is a sine wave, then it is known as sinusoidal alternating voltage, which is given by the relation.
$v=\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}$
Question 10.
An inductor blocks AC but it allows DC. Why? and How?
Answer:
An inductor $\mathrm{L}$ is a closely wound helical coil. The steady DC current flowing through $\mathrm{L}$ produces uniform magnetic field around it and the magnetic flux linked remains constant. Therefore there is no self-induction and self-induced emf (back emf). Since inductor behaves like a resistor, DC flows through an inductor.
The AC flowing through $\mathrm{L}$ produces time-varying magnetic field which in turn induces self- induced emf (back emf). This back emf, according to Lenz's law, opposes any change in the current. Since AC varies both in magnitude and direction, its flow is opposed in L. For an ideal inductor of zero ohmic resistance, the back emf is equal and opposite to the applied emf. Therefore L blocks AC.

Question 11.
A capacitor blocks DC but allows AC. Explain.
Answer:
Capacitive reactance, $\mathrm{X}_{\mathrm{C}}=\frac{1}{\omega C}=\frac{1}{2 \pi f c}$
where, $f=0, \mathrm{X}_{\mathrm{C}}=\infty$
where, $f$ is the frequency of the ac supply. In a de circuit $f=0$. Hence the capacitive reactance has infinite value for de and a finite value for ac. In other words, a capacitor serves as a block for de and offers an easy path to ac.
Question 12.
Why dc ammeter cannot read ac?
Answer:
A dc ammeter cannot read ac because, the average value of ac is zero over a complete cycle.
Question 13.
Write down the applications of series RLC resonant circuit.
Answer:
RLC circuits have many applications like filter circuits, oscillators and voltage multipliers, etc. An important use of series RLC resonant circuits is in the tuning circuits of radio and TV systems. The
signals from many broadcasting stations at different frequencies are available in the air. To receive the signal of a particular station, tuning is done.
Question 14.
What is meant by 'Wattful current'?
Answer:
The component of current $\left(\mathrm{I}_{\mathrm{rms}} \cos \Phi\right)$ which is in phase with the voltage is called active component. The power consumed by this current $=\mathrm{V}_{\mathrm{rms}} \mathrm{I}_{\mathrm{rms}} \cos \Phi$. So that it is also known as 'Wattful' current.

LongAnswer Questions
Question 1.
Derive an expression for Mutual Inductance between two long co-axial solenoids.
Answer:
Mutual inductance between two long co-axial solenoids:
Consider two long co-axial solenoids of same length 1. The length of these solenoids is large when compared to their radii so that the magnetic field produced inside the solenoids is uniform and the fringing effect at the ends may be ignored. Let $\mathrm{A}_1$ and $\mathrm{A}_2$ be the area of cross section of the solenoids with $\mathrm{A}_1$ being greater than $\mathrm{A}_2$. The turn density of these solenoids are $\mathrm{n}_1$ and $\mathrm{n}_2$ respectively.

Let $i_1$ be the current flowing through solenoid 1 , then the magnetic field produced inside it is $B_1=\mu_0 n_1 i_1$
As the field lines of $\vec{B}_1$ are passing through the area bounded by solenoid 2 , the magnetic flux is linked with each turn of solenoid 2 due to solenoid 1 and is given by
$
\phi_{21}=\int_{\mathrm{A}_2} \overrightarrow{\mathrm{B}_1} \cdot d \overrightarrow{\mathrm{A}}=\mathrm{B}_1 \mathrm{~A}_2=\left(\mu_0 n_1 i_1\right) \mathrm{A}_2
$
since $\theta=0^{\circ}$
The flux linkage of solenoid 2 with total turns $\mathrm{N}_2$ is
$
\begin{aligned}
& \mathrm{N}_2 \Phi_{21}=\left(\mathrm{n}_2 1\right)\left(\mu_0 \mathrm{n}_1 \mathrm{i}_1\right) \\
& \text { since } \mathrm{N}_2=\mathrm{n}_2 1 \\
& \mathrm{~N}_2 \Phi_{21}=\left(\mu_0 \mathrm{n}_1 \mathrm{n}_2 \mathrm{~A}_2 \mathrm{l}\right) \mathrm{i}_1 \ldots \ldots \text { (1) }
\end{aligned}
$
From equation of mutual induction
$
\mathrm{N}_2 \Phi_{21}=\mathrm{M}_{21} \mathrm{i}_1 \ldots \ldots(2)
$
Comparing the equations (1) and (2),

$\mathrm{M}_{21}=\mu_0 \mathrm{n}_1 \mathrm{n}_2 \mathrm{~A}_2 1 \ldots . .(3)$
This gives the expression for mutual inductance $M_{21}$ of the solenoid 2 with respect to solenoid 1 .
Similarly, we can find mutual inductance $\mathrm{M}_{21}$ of solenoid 1 with respect to solenoid 2 as given below.
The magnetic field produced by the solenoid 2 when carrying a current $i_2$ is
$\mathrm{B}_2=\mu_0 \mathrm{n}_2 \mathrm{i}_2$
This magnetic field $B_2$ is uniform inside the solenoid 2 but outside the solenoid 2 , it is almost zero.
Therefore for solenoid 1, the area $\mathrm{A}_2$ is the effective area over which the magnetic field $\mathrm{B}_2$ is present; not area $\mathrm{A}_2$ Then the magnetic flux $\Phi_{12}$ linked with each turn of solenoid 1 due to solenoid 2 is
$
\Phi_{12}=\int_{\mathrm{A}_2} \overrightarrow{\mathrm{B}_2} \cdot d \overrightarrow{\mathrm{A}}=\mathrm{B}_2 \mathrm{~A}_2=\left(\mu_0 n_2 i_2\right) \mathrm{A}_2
$
The flux linkage of solenoid 1 with total turns $\mathrm{N}_1$ is
[Since $\left.\mathrm{N}_1=\mathrm{n}_1 1\right]$
[Since $\left.\mathrm{N}_1 \Phi_{12}=\mathrm{M}_{12} \mathrm{i}_2\right]$
$\mathrm{N}_1 \Phi_{12}=\left(\mathrm{n}_1 \mathrm{l}\right)\left(\mu_0 \mathrm{n}_2 \mathrm{i}_2\right) \mathrm{A}_2$
$\mathrm{N}_1 \Phi_{12}=\left(\mu_0 \mathrm{n}_1 \mathrm{n}_2 \mathrm{~A}_2 \mathrm{l}\right) \mathrm{i}_2$
$\mathrm{M}_{12} \mathrm{i}_2=\left(\mu_0 \mathrm{n}_1 \mathrm{n}_2 \mathrm{~A}_2 \mathrm{l}\right) \mathrm{i}_2$
Therefore, we get
$
\therefore \mathrm{M}_{12}=\mu_0 \mathrm{n}_1 \mathrm{n}_2 \mathrm{~A}_2 1 \ldots \ldots . \text { (4) }
$
From equation (3) and (4), we can write
$
\mathrm{M}_{12}=\mathrm{M}_{21}=\mathrm{M} \ldots \ldots . \text { (5) }
$
In general, the mutual inductance between two long co-axial solenoids is given by $\mathrm{M}=\mu_0 \mathrm{n}_1 \mathrm{n}_2 \mathrm{~A}_2 1 \ldots \ldots .(6)$
If a dielectric medium of relative permeability' pr is present inside the solenoids, then $\mathrm{M}=\mu \mathrm{n}_1 \mathrm{n}_2 \mathrm{~A}_2 1$
or $\mathrm{M}=\mu_0 \mu_{\mathrm{r}} \mathrm{n}_1 \mathrm{n}_2 \mathrm{~A}_2 1$
Question 2.
How will you define the unit of mutual-inductance?
Answer:
Unit of mutual inductance:
The unit of mutual inductance is also henry $(\mathrm{H})$.
If $\mathrm{i}_{\mathrm{A}}=1 \mathrm{~A}$ and $\mathrm{N}_2 \Phi_{21}=1 \mathrm{~Wb}$ turns, then $\mathrm{M}_{21}=1 \mathrm{H}$.
Therefore, the mutual inductance between two coils is said to be one henry if a current of $1 \mathrm{~A}$ in coil 1 produces unit flux linkage in coil 2 .
If $\frac{d i_1}{2}=1 \mathrm{As}^{-1}$ and $\varepsilon_2=-1 \mathrm{~V}$, theen $\mathrm{M}_{21}=1 \mathrm{H}$.
Therefore, the mutual inductance between two coils is one henry if a current changing at the rate of 1As-1 in coil 1 induces an opposing emf of IV in coil 2.

Question 3.
Find out the phase relationship between voltage and current in a pure resister circuit.
Answer:
AC circuit containing pure resistor:
Consider a circuit containing a pure resistor of resistance $\mathrm{R}$ connected across an alternating voltage
source. The instantaneous value of the alternating voltage is given by

$v=V_{\mathrm{m}} \sin \omega \mathrm{t} \ldots . .(1)$
An alternating current $i$ flowing in the circuit due to this voltage, develops a potential drop across $R$ and is given by $\mathrm{V}_{\mathrm{R}}=\mathrm{iR} \ldots \ldots(2)$
Kirchoff's loop rule states that the algebraic sum of potential differences in a closed circuit is zero. For this resistive circuit, $v-\mathrm{V}_{\mathrm{R}}=0$
From equation (1) and (2),
$\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t}=\mathrm{iR}$
$\Rightarrow \mathrm{i}=\frac{V_m}{R} \sin \omega \mathrm{t}$
$\mathrm{i}=\mathrm{I}_{\mathrm{m}} \sin \omega \mathrm{t}$

Figure: Phasor diagram and wave diagram for $\mathrm{AC}$ circuit with $\mathrm{R}$
where $\mathrm{V} \frac{V_m}{R}=\mathrm{I}_{\mathrm{m}}$ the peak value of alternating current in the circuit. From equations (1) and and (3), it is clear that the applied voltage and the current are in phase with each other in a resistive circuit. It means that they reach their maxima and minima simultaneously. This is indicated in the phasor diagram. The wave diagram also depicts that current is in phase with the applied voltage.
Question 4.
Find out the phase relationship between voltage and current in a pure capacitor circuit.
Answer:
AC circuit containing only a capacitor:
Consider a circuit containing a capacitor of capacitance $\mathrm{C}$ connected across an alternating voltage source.
The alternating voltage is given by

The alternating voltage is given by

$\mathrm{V}_{\mathrm{m}} \sin \omega \mathrm{t} \ldots \ldots(1)$
Let $\mathrm{q}$ be the instantaneous charge on the capacitor. The emf across the capacitor at that instant is $\frac{q}{C}$. According to Kirchoff's loop rule,.
$
\begin{aligned}
& v=\frac{q}{C}=0 \\
& \Rightarrow \mathrm{CV}_{\mathrm{m}} \sin \omega \mathrm{t}
\end{aligned}
$
By the definition of current,
$
\begin{aligned}
& \mathrm{i}=\frac{d q}{d t}=\frac{d}{d t} \mathrm{CV}_{\mathrm{m}} \frac{d}{d t}(\sin \omega \mathrm{t}) \\
& =\mathrm{CV}_{\mathrm{m}} \sin \omega \mathrm{t} \\
& \text { or } \mathrm{i}=\frac{\frac{\mathrm{v}_m}{1}}{\frac{1}{\mathrm{C}}} \sin \left(\omega t+\frac{\pi}{2}\right) \\
& \mathrm{i}=\mathrm{I}_{\mathrm{m}} \sin \left(\omega t+\frac{\pi}{2}\right) \ldots \ldots(2)
\end{aligned}
$
where $\frac{\frac{\mathrm{V}_m}{\frac{1}{1}}}{\frac{1}{\mathrm{C}}}=\mathrm{I}_{\mathrm{m}}$, the peak value of the alternating current. From equation (1) and (2), it is clear that current leads the applied voltage by $\pi / 2$ in a capacitive circuit. The wave diagram for a capacitive circuit also shows that the current leads the applied voltage by $90^{\circ}$.
Question 5.
What are LC oscillation? and explain the generation of LC oscillation.
Answer:
Whenever energy is given to a circuit containing a pure inductor of inductance $L$ and a capacitor of capacitance $\mathrm{C}$, the energy oscillates back and forth between the magnetic field of the inductor and the electric field of the capacitor. Thus the electrical oscillations of definite frequency are generated. These oscillations are called LC oscillations.
Generation of LC oscillations:
Let us assume that the capacitor is fully charged with maximum charge $\mathrm{Q}_{\mathrm{m}}$ at the initial stage. So that the energy stored in the capacitor is maximum and is given by $\mathrm{U}_{\mathrm{Em}}=\frac{\mathrm{Q}_{\mathrm{m}}^2}{2 \mathrm{C}}$ As there is no current in the inductor, the energy stored in it is zero i.e., UB $=0$. Therefore, the total energy is wholly electrical.
The capacitor now begins to discharge through the inductor that establishes current $\mathrm{i}$ in clockwise direction. This current produces a magnetic field around the inductor and the energy stored in the inductor is given by $\mathrm{U}_{\mathrm{B}}=\frac{L i_2}{2}$. As the charge in the capacitor decreases, the energy stored in it also
decreases and is given by $\mathrm{U}_{\mathrm{E}}=\frac{q_2}{2 C}$. Thus there is a transfer of some part of energy from the capacitor to the inductor. At that instant, the total energy is the sum of electrical and magnetic energies.

When the charges in the capacitor are exhausted, its energy becomes zero i.e., $\mathrm{UE}=0$. The energy is fully transferred to the magnetic field of the inductor and its energy is maximum. This maximum energy is given by $\mathrm{UB}=\frac{\mathrm{LI}_{\mathrm{m}}^2}{2}$ where $\mathrm{I}_{\mathrm{m}}$ is the maximum current flowing in the circuit. The total energy is wholly magnetic.
Even though the charge in the capacitor is zero, the current will continue to flow in the same direction because the inductor will not allow it to stop immediately. The current is made to flow with decreasing magnitude by the collapsing magnetic field of the inductor. As a result of this, the capacitor begins to charge in the opposite direction. A part of the energy is transferred from the inductor back to the capacitor. The total energy is the sum of the electrical and magnetic energies.
When the current in the circuit reduces to zero, the capacitor becomes frilly charged in the opposite direction. The energy stored in the capacitor becomes maximum. Since the current is zero, the energy stored in the inductor is zero. The total energy is wholly electrical. The state of the circuit is similar to the initial state but the difference is that the capacitor is charged in opposite direction. The capacitor then starts to discharge through the inductor with anti-clockwise current. The total energy is the sum of the electrical and magnetic energies.

As already explained, the processes are repeated in opposite direction. Finally, the circuit returns to the initial state. Thus, when the circuit goes through these stages, an alternating current flows in the circuit. As this process is repeated again and again, the electrical oscillations of definite frequency are generated. These are known as LC oscillations. In the ideal LC circuit, there is no loss of energy. Therefore, the oscillations will continue indefinitely. Such oscillations are called undamped oscillations.