Physics

Electrical Machines and Circuits

720 Questions

Electrical machines and circuits involve the principles of alternating and direct current, transformers, diodes, and power consumption. These concepts are essential for engineering and technical exams. Review these practice questions to test your knowledge of circuit analysis.

Transformers and impedanceAlternating and direct currentCircuit components and diodesPower consumption calculations

Electrical Machines and Circuits Questions

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

A source of 220 V is applied in an A C circuit . The value of resistance is 220 $\Omega$. Frequency & inductance are 50Hz & 0.7 H then wattless current is 

  1. 0.5 amp

  2. 0.7 amp

  3. 1.0 amp

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A source= $220V$

The value of resistance= $220 \Omega$
Frequency= $50 Hz$
Inductance= $0.7H$
Find the wattless current= ?
Wattless component of current is $i=i _v\sin \theta$
                                                            $=\cfrac {Ev}{z}\sin \theta$
where, $z=$ impedance of $L-R$ circuit
                $=\sqrt {R^2+L^2W^2}$ so,
$i=\cfrac {220}{\sqrt {R^2+L^2+W^2}}\sin \theta$ from impedance triangle,
$\sin \theta= \cfrac {LW}{\sqrt {R^2+L^2W^2}}$
$\Rightarrow i=\cfrac {220}{\sqrt {R^2+L^2W^2}}\cfrac {LW}{\sqrt {R^2+L^2W^2}}$
        $=\cfrac {220}{R^2+L^2W^2}LW$
        $=\cfrac {220 \times 0.7 \times 2 \Pi \times 50}{(220)^2+(0.7\times 2\Pi \times 50)^2}$
        $=\cfrac {220 \times 220}{(220)^2+(220)^2}$
        $=0.5 A$ .

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

The time constant of a circuit is 10 sec, When a resistance of $ 100 \Omega $ is connected in series in a previous circuit then time constant becomes 2 second,then the self inductance of the circuit is;-

  1. $250 H$
  2. $50H$
  3. $150 H$
  4. $25 H$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In LR circuit,

The time constant $\tau=\dfrac{L}{R}$
$10=\dfrac{L}{R}$
$L=10R$. . . . . . .(1)
When Resistance $100\Omega $ is connect in series, than the time constant is
$\tau'=\dfrac{L}{R+100}=2s$
$L=2R+200$. . . . . . .(2)
Equating equation (1 ) and (2), we get
$2R+200=10R$
$8R=200$
$R=25\Omega$
From equation (1),
$L=10R=10\times 25$
$L=250H$

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

the number of turn of primary and secondary coil of the transformer is 5 and 10 respectively ad the mutual inductance is 25 H. if the number f turns of the primary and secondary is made 10 and 5 , then the mutual inductance of the coils will be

  1. 6.25 H

  2. 12.5 H

  3. 25 H

  4. 50 H
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$M=\mu _o\mu _r \cfrac {N _1N _2}{l}A$

$M \propto N _1N _2$
Since $N _1N _2=10 \times 5= 5 \times 10=50$ in both cases.
Mutual inductance will remain same.

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

if the length and area of cross section of an inductor remain same but the number of turns is doubled its self inductance will become:

  1. half

  2. four time

  3. double

  4. one- fourth
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Self-inductance L is proportional to the square of the total number of turns (N^2). If N is doubled, L becomes 2^2 = 4 times the original.

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

If a spark is produced on removing the load from an AC circuit then the element connected in the circuit is

  1. high resistance

  2. high capacitance

  3. high inductance

  4. high impedance
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

On removal of load from the circuit, the circuit suddenly becomes an open circuit.
Thus $ \dfrac{di}{dt} \rightarrow \infty $
For sparking, high voltage must appear across the open ends. This will happen only in case of an inductor as the voltage drop across the inductor is $ L\dfrac{di}{dt} $
Therefore, the circuit has high inductance.

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

If length of a solenoid is increased then what change should be made on no. of turns to keep self inductance constant-

  1. increase

  2. remain same

  3. decrease

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$N$=total number of turns in solenoid
$l$=length of solenoid
$R$=radius of cross section of solenoid
Magnetic field inside solenoid =$B=\mu _o \dfrac{N}{l}i$  where, $\dfrac{N}{l}$=number of turns per unit length

Now, emf induced= $E=NBA$

$\implies E=N\times \mu _o\dfrac{N}{l}i\times \pi R^2$

$\implies E=\dfrac{\mu _{o}N^2\pi R^2}{l}i=Li$
                                                                                    where $L$=self inductance
                                      
Thus, $L=\dfrac{\mu _o N^2 \pi R^2}{l}$

For constant $L$,    $N^2\propto l$

Hence, on increasing length of coil, number of turns should be increased to keep $L$ constant.

Answer-(A)
Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

Reactance of a coil is $157\Omega$. On connecting the coil across a source of frequency $ 100Hz$, the current lags behind e.m.f. by ${ 45 }^{ o }$. The inductance of the coil is _________.

  1. $0.25 H$
  2. $0.5 H$
  3. $4H$
  4. $314 H$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since the phase angle is $45^{\circ}$,

$\dfrac{X _L}{R}=tan\phi=tan45^{circ}=1$
$\implies X _L=R$
$\implies \omega L=R$
$\implies 2\pi f L=R$
$\implies L=\dfrac{R}{2\pi f}$
$=\dfrac{157}{2\pi\times 100}H$
$=0.25H$

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

The electrical analog of mass is

  1. Diode

  2. Capacitance

  3. Inductance

  4. Resistance
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As per mechanical-electrical analog, displacement is analogous to charge and force analogous to voltage.

From newton's second law of motion,
$F = m \cfrac{d^2x}{dt^2}$

For a diode, voltage and current are exponentially related and is non-linear.
For capacitance,  $V = \cfrac{Q}{C}$
For inductance, $V = L\cfrac{dI}{dt} = L\cfrac{d^2 q}{dt^2}$
For resistance, $V = IR = R \cfrac{dq}{dt}$

By comparing the above equations, it can be concluded that electrical analog of mass is inductance. 

Multiple choice introduction to induction electromagnetic induction electromagnetic induction and alternating currents physics

Find the necessary inductance. if 110 V, 10 W rating bulb is to be used with 220 V A.C source having frequency 50 Hz.

  1. L=8.90 H

  2. L=6.75 H

  3. L=7.25 H

  4. L=6.5 H
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $V=110v$ ,$P=10W$ ,$V _0=220 w$ ,$f=50Hz$

Current through series inductor,
Current through bulb=$\dfrac{P}{V}=\dfrac{10W}{110V}=0.09A$
Voltage across inductor,$V _{ind}=\sqrt{V _0^{2} -V^{2}}$=$\sqrt{220^{2}-110^2}=191V$
Reactance of inductor,$R=\dfrac{V _{ind}}{I}=\dfrac{191}{0.09}=2122.22$
Also,$R=2 \pi fL$ or $L$=$\dfrac{R}{2 \pi f}$=$\dfrac{2122.22}{2 \pi 50}$=$6.75H$

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

A parallel plate condenser of capacity $10\mu F$ is connected to an A.C. supply voltage $e = 4\sin \left( {100\pi t} \right)$. The maximum displacement current:

  1. $4\pi \mu A$
  2. $4\pi mA$
  3. $2.8\pi \mu A$
  4. $2\pi \mu A$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Displacement current, ${i _d} = {\varepsilon _ \circ }A\frac{{dE}}{{dt}}$
where, $A$ is area and $E$ is electric field
$\begin{array}{l}E = \frac{e}{d} = \frac{{4\sin (100\pi t)}}{d}\{i _d} = {\varepsilon _ \circ }A \times \frac{1}{d}\left[ {\frac{d}{{dt}}\left( {4\sin 100\pi t} \right)} \right]\ = \frac{{{\varepsilon _ \circ }A}}{d}.4(cos100\pi t) \times 100\pi \ = C.400\pi .cos100\pi t\{\left( {{i _d}} \right) _{\max }} = \left( {10 \times {{10}^{ - 6}}} \right) \times 400\pi A\ = 4\pi  \times {10^{ - 3}}A = 4\pi mA\end{array}$

Multiple choice physics semiconductors band theory of solids, a brief introduction electron energies in solids energy bands

A transformer has 500 turns in its primary and 1000 turns in its secondary winding.The primary voltage is 200 V and the load in the secondary is 100 ohm.Calculate the current in the primary,assuming it to be a ideal transformer.

  1. 25 A

  2. 45 A

  3. 8A

  4. 22 A
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In usual rotation we have -

$\Rightarrow \dfrac{V _s}{V _p}=\dfrac{N _s}{N _p}$
$\therefore V _S=V _p\times \dfrac{N _s}{N _p}$
          $=200v\times \dfrac{1000}{500}$
          $=400V$
There is load resistance of $100\Omega$ in the secondary circuit. Therefore current in the secondary 
$\Rightarrow i _s=\dfrac{V _s}{R _s}=\dfrac{400V}{100\Omega}=4A$
In an ideal transformer the output power and input power are equal that is 
$\Rightarrow V _s\times i _s=V _p\times i _p$
$\therefore i _p=i _s \times \dfrac{V _s}{V _p}=4A\times \dfrac{400V}{200V}$
                            $=4A\times 2$
                            $=8A$
Hence, the answer is $8A.$