MCQ 511 Mark
Which of the following is constructed on the principle of electromagnetic induction
AnswerIn a generator. e.m.f. is induced according as Lenz's rule.
View full question & answer→MCQ 521 Mark
A solenoid of length $/$ metre has self $-$ inductance $L$ henry. If number of turns are doubled, its self inductance
AnswerCorrect option: C. Becomes $4 L$ henry
Becomes $4 L$ henry
View full question & answer→MCQ 531 Mark
An inductor of $2$ henry and a resistance of $10 \mathrm{ohms}$ are connected in series with a battery of $5$ volts. The initial rate of change of current is
- A
$0.5 \mathrm{amp} / \mathrm{sec}$
- B
$2.0 \mathrm{amp} / \mathrm{sec}$
- ✓
$2.5 \mathrm{amp} / \mathrm{sec}$
- D
$0.25 \mathrm{amp} / \mathrm{sec}$
AnswerCorrect option: C. $2.5 \mathrm{amp} / \mathrm{sec}$
$ i=i_0\left(1-e^{\frac{-R t}{L}}\right) \Rightarrow \frac{d i}{d t}=\frac{d}{d t} i_0-\frac{d}{d t} i_0 e^{\frac{-R t}{L}} $
$ \Rightarrow \frac{d i}{d t}=0-i_0\left(-\frac{R}{L}\right) e^{-\frac{R t}{L}}=\frac{i_0 R}{L} e^{-\frac{R t}{L}}$
Initially, $t=0 \Rightarrow \frac{d i}{d t}=\frac{i_0 \times R}{L=\frac{E}{L}}=\frac{5}{2}=2.5 \mathrm{amp} / \mathrm{sec}$.
View full question & answer→MCQ 541 Mark
The equivalent inductance of two inductances is $2.4$ henry when connected in parallel and $10$ henry when connected in series. The difference between the two inductances is
- ✓
$2$ henry
- B
$3$ henry
- C
$4$ henry
- D
$5$ henry
AnswerCorrect option: A. $2$ henry
We know that In series
$L_{e q}=L_1+L_2$
$10=L_1+L_2 $
$L_1=10-L_2$
In parallel
$L_{e q}=\frac{L_1+L_2}{L_1+L_2}$
$\frac{L_1 L_2}{10}=2.4$
$L_1 L_2=24 $
$\left(10-L_2\right) / L_2=24$
$L_2^2-10 L_2+24=0$
$L_2^2-6 L_2-4 L_2+24=0 $
$L_2\left(L_2-6\right)-4\left(L_2-6\right)=0$
$\left(L_2-4\right)\left(L_2-6\right)=0$
$L_2=4 \text { or } L_2=6$
View full question & answer→MCQ 551 Mark
An e.m.f. of $12$ volts is induced in a given coil when the current in it changes at the rate of $48$ amperes per minute. The self inductance of the coil is
- A
$0.25$ henry
- ✓
$15$ henry
- C
$1.5$ henry
- D
$9.6$ henry
AnswerCorrect option: B. $15$ henry
(b) $L=\frac{e}{d i / d t}=\frac{12}{48 / 60}=15 H$
View full question & answer→MCQ 561 Mark
The armature of dc motor has $20 \Omega$ resistance. It draws current of $1.5$ ampere when run by $220$ volts dc supply. The value of back e.m.f. induced in it will be
- A
$150 \mathrm{~V}$
- B
$170 \mathrm{~V}$
- C
$180 \mathrm{~V}$
- ✓
$190 \mathrm{~V}$
AnswerCorrect option: D. $190 \mathrm{~V}$
(d) $i=\frac{E-e}{R} \Rightarrow 1.5=\frac{220-e}{20} \Rightarrow e=190 \mathrm{~V}$
View full question & answer→MCQ 571 Mark
The mutual inductance between a primary and secondary circuits is $0.5 \mathrm{H}$. The resistances of the primary and the secondary circuits are $20\ \mathrm{ohms}$ and $5\ \mathrm{ohms}$ respectively. To generate a current of $0.4 \mathrm{~A}$ in the secondary, current in the primary must be changed at the rate of
- ✓
$4.0 \mathrm{~A} / \mathrm{s}$
- B
$16.0 \mathrm{~A} / \mathrm{s}$
- C
$1.6 \mathrm{~A} / \mathrm{s}$
- D
$8.0 \mathrm{~A} / \mathrm{s}$
AnswerCorrect option: A. $4.0 \mathrm{~A} / \mathrm{s}$
$e_2=M \frac{d i_1}{d t} \Rightarrow i_2 R_2=M \frac{d i_1}{d t} \Rightarrow 0.4 \times 5=0.5 \times \frac{d i_1}{d t} $
$ \Rightarrow \frac{d i_1}{d t}=4 \mathrm{~A} / \mathrm{sec} .$
View full question & answer→MCQ 581 Mark
The self inductance of a coil is $L$. Keeping the length and area same, the number of turns in the coil is increased to four times. The self inductance of the coil will now be
- A
$\frac{1}{4} L$
- B
$L$
- C
$4 L$
- ✓
$16 \mathrm{~L}$
AnswerCorrect option: D. $16 \mathrm{~L}$
$16 \mathrm{~L}$
View full question & answer→MCQ 591 Mark
A long horizontal metallic rod with length along the east$-$west direction is falling under gravity. The potential difference between its two ends will
View full question & answer→MCQ 601 Mark
In a circuit with a coil of resistance $2 \mathrm{ohms}$, the magnetic flux changes from $2.0 \mathrm{~Wb}$ to $10.0 \mathrm{~Wb}$ in $0.2$ second. The charge that flows in the coil during this time is
- A
$5.0$ coulomb
- ✓
$4.0$ coulomb
- C
$1.0$ coulomb
- D
$0.8$ coulomb
AnswerCorrect option: B. $4.0$ coulomb
$\Delta Q=\frac{\Delta \phi}{R}=\frac{(10-2)}{2}=4 C$
View full question & answer→MCQ 611 Mark
A coil of $100$ turns and area $5$ square centimetre is placed in a magnetic field $\mathrm{B}=0.2 \mathrm{~T}$. The normal to the plane of the coil makes an angle of $60^{\circ}$ with the direction of the magnetic field. The magnetic flux linked with the coil is
AnswerCorrect option: A. $5 \times 10^{-3} \mathrm{~Wb}$
$ \phi=N B A \cos \theta=100 \times 0.2 \times 5 \times 10^{-4} \cos 60^{\circ}$
$ =5 \times 10^{-3} \mathrm{~Wb}$
View full question & answer→MCQ 621 Mark
In a transformer the primary has $500$ turns and secondary has $50$ turns. $100$ volts are applied to the primary coil, the voltage developed in the secondary will be
- A
$1 \mathrm{~V}$
- B
$10 \mathrm{~V}$
- ✓
$1000 \mathrm{~V}$
- D
$10000 \mathrm{~V}$
AnswerCorrect option: C. $1000 \mathrm{~V}$
View full question & answer→MCQ 631 Mark
A coil of self inductance $50$ henry is joined to the terminals of a battery of e.m.f. $2$ volts through a resistance of $10 \mathrm{ohm}$ and a steady current is flowing through the circuit. If the battery is now disconnected, the time in which the current will decay to $1 / e$ of its steady value is
- A
$500$ seconds
- B
$50$ seconds
- ✓
$5$ seconds
- D
$0.5$ seconds
AnswerCorrect option: C. $5$ seconds
(c) Time in which the current will decay to $\frac{1}{e}$ of its steady value is $t=\tau=\frac{L}{R}=\frac{50}{10}=5$ seconds
View full question & answer→MCQ 641 Mark
In a coil of self inductance $0.5$ henry, the current varies at a constant rate from zero to $10$ amperes in $2$ seconds. The e.m.f. generated in the coil is
- A
$10$ volts
- B
$5$ volts
- ✓
$2.5$ volts
- D
$1.25$ volts
AnswerCorrect option: C. $2.5$ volts
(c) $\frac{\Delta i}{\Delta t}=\frac{10}{2}=5 \mathrm{~A} / \mathrm{sec} \Rightarrow e=L \frac{\Delta i}{\Delta t}=0.5 \times 5=2.5$ volts
View full question & answer→MCQ 651 Mark
A loss free transformer has $500$ turns on its primary winding and $2500$ in secondary. The meters of the secondary indicate $200 \mathrm{volts}$ at $8$ amperes under these conditions. The voltage and current in the primary is
- A
$100 \mathrm{~V}, 16 \mathrm{~A}$
- ✓
$40 \mathrm{~V}, 40 \mathrm{~A}$
- C
$160 \mathrm{~V}, 10 \mathrm{~A}$
- D
$80 \mathrm{~V}, 20 \mathrm{~A}$
AnswerCorrect option: B. $40 \mathrm{~V}, 40 \mathrm{~A}$
(b) $\frac{V_p}{V_s}=\frac{N_p}{N_s}=\frac{500}{2500}=\frac{1}{5} \Rightarrow V_p=\frac{200}{5}=40 \mathrm{~V}$
Also $i_p V_p=i_s V_s \Rightarrow i_p=i_s \frac{V_s}{V_p}=8 \times 5=40 \mathrm{~A}$
View full question & answer→MCQ 661 Mark
Two identical coaxial circular loops carry current each circulating in the clockwise direction. If the loops are approaching each other, then
AnswerCorrect option: C. Current in each loop decreases
Current in each loop decreases
View full question & answer→MCQ 671 Mark
The inductance of a coil is $60 \mu \mathrm{H}$. A current in this coil increases from $1.0 A$ to $1.5 A$ in $0.1$ second. The magnitude of the induced e.m.f. is
- A
$60 \times 10^{-6} \mathrm{~V}$
- B
$300 \times 10^{-4} \mathrm{~V}$
- C
$30 \times 10^{-4} \mathrm{~V}$
- ✓
$3 \times 10^{-4} \mathrm{~V}$
AnswerCorrect option: D. $3 \times 10^{-4} \mathrm{~V}$
(d) $e=L \frac{d i}{d t}=60 \times 10^{-6} \cdot \frac{(1.5-1.0)}{0.1}=3 \times 10^{-4}$ volt
View full question & answer→MCQ 681 Mark
A transformer connected to $220$ volt line shows an output of $2 A$ at 1$1000$ volt. The efficiency is $100 \%$. The current drawn from the line is
- ✓
$100 \mathrm{~A}$
- B
$200 \mathrm{~A}$
- C
$22 \mathrm{~A}$
- D
$11 \mathrm{~A}$
AnswerCorrect option: A. $100 \mathrm{~A}$
$100 \mathrm{~A}$
View full question & answer→MCQ 691 Mark
Two conducting circular loops of radii $R_1$ and $R_2$ are placed in the same plane with their centres coinciding. If $R_1 \gg R_2$, the mutual inductance $M$ between them will be directly proportional to
- A
$R_1 / R_2$
- B
$R_2 / R_1$
- C
$R_1^2 / R_2$
- ✓
$R_2^2 / R_1$
AnswerCorrect option: D. $R_2^2 / R_1$
$R_2^2 / R_1$
View full question & answer→MCQ 701 Mark
The unit of magnetic flux is
MCQ 711 Mark
The dimensions of magnetic flux are
- A
$M L T^{-2} A^{-2}$
- B
$M L^2 T^{-2} A^{-2}$
- C
$M L^2 T^{-1} A^{-2}$
- ✓
$M L^2 T^{-2} A^{-1}$
AnswerCorrect option: D. $M L^2 T^{-2} A^{-1}$
View full question & answer→MCQ 721 Mark
The unit of inductance is
Answer(c) $e=L \frac{d i}{d t} \Rightarrow L=\frac{\text { Volt }-\mathrm{sec}}{\text { Ampere }}$
View full question & answer→MCQ 731 Mark
An inductance $L$ and a resistance $R$ are first connected to a battery. After some time the battery is disconnected but $L$ and $R$ remain connected in a closed circuit. Then the current reduces to $37 \%$ of its initial value in
- ✓
(a) $R L \mathrm{sec}$
- B
(b) $\frac{R}{L} \sec$
- C
(c) $\frac{L}{R} \sec$
- D
(d) $\frac{1}{L R} \sec$
AnswerCorrect option: A. (a) $R L \mathrm{sec}$
(a) Current at any instant of time $t$ after closing an $L-R$ circuit is given by $I=I_0\left[1-e^{\frac{-R}{L} t}\right]$Time constant $t=\frac{L}{R}$$\begin{aligned}& \therefore I=I_0\left[1-e^{\frac{-R}{L} \times \frac{L}{R}}\right]=I_0\left(1-e^{-1}\right)=I_0\left(1-\frac{1}{e}\right) \\& =I_0\left(1-\frac{1}{2.718}\right)=0.63 I_0=63 \% \text { of } I_0\end{aligned}$
View full question & answer→MCQ 741 Mark
The current flowing in a coil of self inductance $0.4 \mathrm{mH}$ is increased by $250 \mathrm{~mA}$ in $0.1 \mathrm{sec}$. The e.m.f. induced will be
- A
$+1 \mathrm{~V}$
- B
$-1 \mathrm{~V}$
- C
$+1 \mathrm{mV}$
- ✓
$-1 \mathrm{mV}$
AnswerCorrect option: D. $-1 \mathrm{mV}$
(d) $e=-L \frac{d i}{d t}=-0.4 \times 10^{-3} \times \frac{250 \times 10^{-3}}{0.1}=-1 \mathrm{mV}$
View full question & answer→MCQ 751 Mark
A conducting rod of length $l$ is falling with a velocity $v$ perpendicular to a uniform horizontal magnetic field $B$. The potential difference between its two ends will be
- A
$2 B / v$
- ✓
$B / v$
- C
$\frac{1}{2} B l v$
- D
$B^2 l^2 v^2$
AnswerCorrect option: B. $B / v$
View full question & answer→MCQ 761 Mark
A coil of area $80$ square $\mathrm{cm}$ and $50$ turns is rotating with $2000$ revolutions per minute about an axis perpendicular to a magnetic field of $0.05$ Tesla. The maximum value of the e.m.f. developed in it is
- A
$200 \pi$ volt
- B
$\frac{10 \pi}{3}$ volt
- ✓
$\frac{4 \pi}{3}$ volt
- D
$\frac{2}{3}$ volt
AnswerCorrect option: C. $\frac{4 \pi}{3}$ volt
$e=N B A \omega ; \omega=2 \pi f=2 \pi \times \frac{2000}{60}$
$ \therefore e=50 \times 0.05 \times 80 \times 10^{-4} \times 2 \pi \times \frac{2000}{60}=\frac{4 \pi}{3}$
View full question & answer→MCQ 771 Mark
- A
The magnitude of the induced e.m.f.
- ✓
The direction of the induced current
- C
Both the magnitude and direction of the induced current
- D
The magnitude of the induced current
AnswerCorrect option: B. The direction of the induced current
View full question & answer→MCQ 781 Mark
The north pole of a long horizontal bar magnet is being brought closer to a vertical conducting plane along the perpendicular direction. The direction of the induced current in the conducting plane will be
MCQ 791 Mark
An e.m.f. of $15$ volt is applied in a circuit containing $5$ henry inductance and $10 \mathrm{ohm}$ resistance. The ratio of the currents at time $t=\infty$ and at $t=1$ second is
AnswerCorrect option: B. $\frac{e^2}{e^2-1}$
(b) $i=i_0\left(1-e^{-R t / L}\right)$$i_0=\frac{E}{R}$ (Steady current) when $t=\infty$
$ i_{\infty}=\frac{E}{R}\left(1-e^{-\infty}\right)=\frac{5}{10}=1.5$
$ i_1=1.5\left(1-e^{-R / L}\right)=1.5\left(1-e^{-2}\right) \Rightarrow \frac{i_{\infty}}{i_1}=\frac{1}{1-e^{-2}}=\frac{e^2}{e^2-1}$
View full question & answer→MCQ 801 Mark
In a lossless transformer an alternating current of $2$ amp is flowing in the primary coil. The number of turns in the primary and secondary coils are $100$ and $20$ respectively. The value of the current in the secondary coil is
- A
$0.08 \mathrm{~A}$
- B
$0.4 \mathrm{~A}$
- C
$5 \mathrm{~A}$
- ✓
$10 \mathrm{~A}$
AnswerCorrect option: D. $10 \mathrm{~A}$
$10 \mathrm{~A}$
View full question & answer→MCQ 811 Mark
The pointer of a dead-beat galvanometer gives a steady deflection because
- ✓
Eddy currents are produced in the conducting frame over which the coil is wound
- B
Its magnet is very strong
- C
Its pointer is very light
- D
Its frame is made of abonite
AnswerCorrect option: A. Eddy currents are produced in the conducting frame over which the coil is wound
View full question & answer→MCQ 821 Mark
Self induction of a solenoid is
AnswerCorrect option: C. Directly proportional to area of cross$-$section
Directly proportional to area of cross$-$section
View full question & answer→MCQ 831 Mark
A copper rod of length $l$ is rotated about one end perpendicular to the magnetic field $B$ with constant angular velocity $\omega$. The induced e.m.f. between the two ends is
AnswerCorrect option: A. $\frac{1}{2} B \omega l^2$
View full question & answer→MCQ 841 Mark
A metal conductor of length $1 \mathrm{~m}$ rotates vertically about one of its ends at angular velocity $5$ radians per second. If the horizontal component of earth's magnetic field is $0.2 \times 10^{-4} \mathrm{~T}$, then the e.m.f. developed between the two ends of the conductor is
AnswerCorrect option: D. $50 \mu \mathrm{V}$
(d) $e=\frac{1}{2} B \omega r^2=\frac{1}{2} \times 0.2 \times 10^{-4} \times 5 \times(1)^2=50 \mu \mathrm{V}$
View full question & answer→MCQ 851 Mark
A transformer is employed to reduce $220 V$ to $1 V$. The primary draws a current of $5 A$ and the secondary $90 A$. The efficiency of the transformer is
- A
(a) $20 \%$
- ✓
(b) $40 \%$
- C
(c) $70 \%$
- D
(d) $90 \%$
AnswerCorrect option: B. (b) $40 \%$
(b) $\frac{N_s}{N_p}=\frac{V_s}{V_p}=\frac{22000}{220}=100$
View full question & answer→MCQ 861 Mark
An ideal transformer has $100$ turns in the primary and $250$ turns in the secondary. The peak value of the ac is $28 \mathrm{~V}$. The r.m.s. secondary voltage is nearest to
- ✓
$50 \mathrm{~V}$
- B
$70 \mathrm{~V}$
- C
$100 \mathrm{~V}$
- D
$40 \mathrm{~V}$
AnswerCorrect option: A. $50 \mathrm{~V}$
$50 \mathrm{~V}$
View full question & answer→MCQ 871 Mark
A coil of area $100 \mathrm{~cm}^2$ has $500$ turns. Magnetic field of $0.1$ weber / metre ${ }^2$ is perpendicular to the coil. The field is reduced to zero in $0.1$ second. The induced e.m.f. in the coil is
- A
$1 \mathrm{~V}$
- ✓
$5 \mathrm{~V}$
- C
$50 \mathrm{~V}$
- D
$0$
AnswerCorrect option: B. $5 \mathrm{~V}$
$ e=-\frac{N\left(B_2-B_1\right) A \cos \theta}{\Delta t}$
$ =-\frac{500 \times(0-0.1) \times 100 \times 10^{-4} \cos 0}{0.1}=5 \mathrm{~V}$
View full question & answer→MCQ 881 Mark
The primary winding of a transformer has $100$ turns and its secondary winding has $200$ turns. The primary is connected to an ac supply of $120 \mathrm{~V}$ and the current flowing in it is $10 \mathrm{~A}$. The voltage and the current in the secondary are
- ✓
$240 \mathrm{~V}, 5 \mathrm{~A}$
- B
$240 V, 10 \mathrm{~A}$
- C
$60 \mathrm{~V}, 20 \mathrm{~A}$
- D
$120 \mathrm{~V}, 20 \mathrm{~A}$
AnswerCorrect option: A. $240 \mathrm{~V}, 5 \mathrm{~A}$
(a) $\frac{N_s}{N_p}=\frac{V_s}{V_p} \Rightarrow \frac{200}{100}=\frac{V_s}{120} \Rightarrow V_s=240 \mathrm{~V}$
also $\frac{V_s}{V_p}=\frac{i_p}{i_s} \Rightarrow \frac{240}{120}=\frac{10}{i_s} \Rightarrow i_s=5 \mathrm{~A}$
View full question & answer→MCQ 891 Mark
If rotational velocity of a dynamo armature is doubled, then induced e.m.f. will become
MCQ 901 Mark
When a wire loop is rotated in a magnetic field, the direction of induced e.m.f. changes once in each
- A
$\frac{1}{4}$ revolution
- ✓
$\frac{1}{2}$ revolution
- C
$1$ revolution
- D
$2$ revolution
AnswerCorrect option: B. $\frac{1}{2}$ revolution
$\frac{1}{2}$ revolution
View full question & answer→MCQ 911 Mark
A coil and a bulb are connected in series with a dc source, a soft iron core is then inserted in the coil. Then
- A
Intensity of the bulb remains the same
- ✓
Intensity of the bulb decreases
- C
Intensity of the bulb increases
- D
AnswerCorrect option: B. Intensity of the bulb decreases
There will be self induction effect when soft iron core is inserted.
View full question & answer→MCQ 921 Mark
A square coil $10^{-2} \mathrm{~m}^2$ area is placed perpendicular to a uniform magnetic field of intensity $10^3 \mathrm{~Wb} / \mathrm{m}^2$. The magnetic flux through the coil is
- ✓
$10$ weber
- B
$10^{-5}$ weber
- C
$10^5$ weber
- D
$100$ weber
AnswerCorrect option: A. $10$ weber
(a) $\phi=B A=10$ weber
View full question & answer→MCQ 931 Mark
A coil of wire of a certain radius has $600$ turns and a self inductance of $108 \mathrm{mH}$. The self inductance of a $2$ similar coil of $500$ turns will be
- A
$74 \mathrm{mH}$
- ✓
$75 \mathrm{mH}$
- C
$76 \mathrm{mH}$
- D
$77 \mathrm{mH}$
AnswerCorrect option: B. $75 \mathrm{mH}$
(b) $\frac{L_B}{L_A}=\left(\frac{n_B}{n_A}\right)^2 \Rightarrow L_B=\left(\frac{500}{600}\right)^2 \times 108=75 \mathrm{mH}$
View full question & answer→MCQ 941 Mark
In a transformer, the coefficient of mutual inductance between the primary and the secondary coil is $0.2$ henry. When the current changes by $5$ ampere/second in the primary, the induced e.m.f. in the secondary will be
- A
$5 \mathrm{~V}$
- ✓
$1 \mathrm{~V}$
- C
$25 \mathrm{~V}$
- D
$10 \mathrm{~V}$
AnswerCorrect option: B. $1 \mathrm{~V}$
(b) $e=M \frac{d i}{d t}=0.2 \times 5=1 \mathrm{~V}$
View full question & answer→MCQ 951 Mark
When the number of turns in a coil is doubled without any change in the length of the coil, its self inductance becomes
Answer$L \propto N^2$ i.e. $\frac{L_1}{L_2}=\left(\frac{N_1}{N_2}\right)^2 \Rightarrow L_2=L_1\left(\frac{N_2}{N_1}\right)^2=4 L_1$
View full question & answer→MCQ 961 Mark
The ratio of secondary to the primary turns in a transformer is $3: 2$. If the power output be $P$, then the input power neglecting all loses must be equal to
- A
$5 P$
- B
$1.5 P$
- ✓
$P$
- D
$\frac{2}{5} P$
View full question & answer→MCQ 971 Mark
A $50 \mathrm{mH}$ coil carries a current of $2$ ampere. The energy stored in joules is
AnswerEnergy stored $E=\frac{1}{2} L i^2=\frac{1}{2} \times 50 \times 10^{-3} \times 4=0.1 \mathrm{~J}$
View full question & answer→MCQ 981 Mark
Average energy stored in a pure inductance $L$ when a current $i$ flows through it, is
- A
$L i^2$
- B
$(b) $2 L i^2$
- C
$\frac{L i^2}{4}$
- ✓
$\frac{L i^2}{2}$
AnswerCorrect option: D. $\frac{L i^2}{2}$
As we know $e=-\frac{d \phi}{d t}=-L \frac{d i}{d t}$Work done against back e.m.f. $e$ in time $d t$ and current $i$ is$d W=-e i d t=L \frac{d i}{d t} i d t=L i d i \Rightarrow W=L \int_0^i i d i=\frac{1}{2} L i^2$
View full question & answer→MCQ 991 Mark
The back e.m.f. induced in a coil, when current changes from 1 ampere to zero in one milli-second, is 4 volts, the self inductance of the coil is
AnswerCorrect option: D. $4 \times 10^{-3} \mathrm{H}$
$e=-L \frac{d i}{d t} \text { but } e=4 V \text { and } \frac{d i}{d t}=\frac{0-1}{10^{-3}}=-1 / 10^{-3}$
$ \therefore \frac{-1}{10^{-3}}(-L)=4 \Rightarrow L=4 \times 10^{-3} \text { henry }$
View full question & answer→MCQ 1001 Mark
In a transformer $220$ ac voltage is increased to $2200$ volts. If the number of turns in the secondary are $2000,$ then the number of turns in the primary will be
Answer$\frac{V_p}{V_s}=\frac{N_p}{N_s} \Rightarrow N_p=\left(\frac{220}{2200}\right) 2000=200$
View full question & answer→