**1) A half wave rectifier has resistive load 500 Ω and
rectifier-alternating voltage of 65V, diode forward resistance is 50 Ω.
Calculate the peak value of current?**

**Given: –**

Vrms = 65V

RL = 500 Ω

RF = 50 Ω

Solutions: –

IM = VM / [RL + RF]

Vrms = VM / √2 => VM = Vrms * √2

IM = 65 X √2 / [500+50]

**Answer: IM = 0.167A**

**2) What is the value of Z in series circuit where XL=9 Ω
XC=5 Ω and R=8 Ω?**

**Solutions: –**

Z= √ [R2 +(XL-XC)2]

Z= √ [82 +(9-5)2]

**Answer: 8.944 Ω**

**3) Sine wave rms value 100V riding on 50 Vdc. Calculate
maximum value of the resulting waveform?**

**Given: –**

Vrms = 100V

Vdc = 50V

Solutions: –

Vrms = VM / √2

VM = 100 * √2

VM = 141.42V

Resulting waveform = VM + Vdc = 141.42 + 50 = 191.42V

**Answer: Resulting waveform = 191.42V**

**4) The conductive loop of the rotor of a simple two poles,
single phase generators at 400 rps. Calculate freq induced output voltage.**

**Given: –**

n = 400 rps = 400 * 60 rpm

p = 2

**Solutions: –**

n = 120 f / p

f = 400 * 60 * 2 / 120 = 400 Hz

**Answer: f = 400 Hz**

**5) Calculate the avg value of 12V peak sine wave for one
complete cycle.**

**Given: –**

VM = 12V

**Solutions: –**

Vavg = 0.637 * VM

Vavg = 0.637 * 12 = 7.644V

**Answer: Vavg = 7.644V**

**Related Formulas: Peak to Peak = VM + VM**

**6) An insulation resistance reading is taken at 20o C found
to be 10 MΩ. What should be expected reading at 40o C?**

Note: The insulation resistance readings are divided by two for every increasing 10o C

The insulation resistance readings are multiplied by two for every decreasing 10o C

**Solutions: –**

At 30o C IR is 10/2 = 5 MΩ

At 40o C IR is 5/2 = 2.5 MΩ

**Answer: 2.5 MΩ**

**7) Convert 0.06 KV into ______ mV**

**Solutions: –**

0.06 * 1000 = 60V

60 * 1000 = 60000 mV

**Answer: 60000 mV**

**8) A fan is drawing 16KW at 800rpm. If the speed is reduced
to 600 rpm the power drawn by the fan would be?**

**Given: –**

P1 = 16 KW

N1 = 800

N2 = 600

**Solutions: –**

[P1/P2] = [N1/N2]3

[16/P2] = [800/600]3

P2 = 6.75 KW

**Answer: P2 = 6.75 KW**

**9) What is the maximum allowable primary current of 2 KVA
step down transformer with a four to one turns ratio is connected across a
440-volt line?**

**Solutions: –**

V1/V2 = N1/N2 = I2/I1 = K (Transformer Ratio)

440/V2 = 4/1

V2 = 110V

The transformer rating in volt-amp = V2 * I2

2000 = 110 * I2

Secondary Current I2 = 18.18 A

I2/I1 = N1/N2

18.18/I1 = 4/1

I1 = 4.545 A

**Answer: Primary current I1 = 4.545 A**

**10) If an AC motor is started and produces 25 horse power,
the KW meter reading will increase by**

**Solutions: –**

1HP = 0.746 KW

25 HP = 0.746 * 25 = 18.65 KW

**Answer: 18.65 KW**

**11) A three-phase alternator is operating at 450 volts with
the switchboard ammeter indicating 300 amps. The kw meter currently indicates
163.6 KW, with a power factor of 0.7. If the power factor increases to 0.8, the
KW meter reading would increase by ____________.**

**Given: –**

V = 450V

I = 300A

Cos ø = 0.8

**Solutions:**

Power = √3 * V * I * Cos ø / 1000

= √3 * 450 * 300 * 0.8 / 1000

= 187.06 KW

Increased power = 187.06 – 163.6 = 23.4KW

**Answer: 23.4KW**

**12) A turbocharger has a rated output of 1200 KW at 60 Hz,
with a no load frequency 61.5 Hz. What is speed drop?**

**Solutions: –**

S.D = (No load speed – Full load speed) / No load speed

= (61.5 – 60) / 61.5 = 0.02439

**Answer: S.D = 0.02439**

**13) A certain appliances 750 W. if it is allowed to run
continuously for 15 days how many KW-hours of energy does it consumes?**

**Solutions: –**

Current consumes per hour = 750 W

For 15 days = 15 * 24 * 750 = 270000 W-h = 270KWh

**Answer: 270KWh**

**14) If it takes 84J of energy to move a charge of 15C from
one point to another. What is the voltage between two points?**

**Given: –**

W = 84

Q = 15

**Solutions: –**

V = W/ Q = 84 / 15 = 5.60V

**Answer: V = 5.60V**

**15) If three resistors of resistance of 30 Ω, 50 Ω, 20 Ω
are connected in series and an output voltage of 20V is applied to this
circuit. Calculate the voltage drop across the 50 Ω resistor?**

**Given: –**

R1 = 30 Ω

R2 = 50 Ω

R3 = 20 Ω

V = 20V

**Solutions: –**

R = R1 + R2 + R3 = 30 + 50 + 20 = 100 Ω

V = IR

I = V / R = 20 / 100 = 0.2A ]

Voltage drop on R2 = I * R2

= 0.2 * 50 = 10V

**Answer: Voltage drop on R2 = 10V**

**16) If resistance of any coil is 120 Ω. Calculate the
resistance when its length is increased by 20% and its dia is increased by 50%.**

**Given: –**

R1 = 120 Ω

L2 = 1.2 L1

D2 = 1.5D1

**Solutions: –**

R = ΡL/A9

R2 = R1 * (A1 * L2) / (A2 * L1) = 120 * {( ) * 1.2 L1} / {( ) * L1}

= 120 * 1.2 / (1.5)2 = 64 Ω

**Answer: R2 = 64Ω**

**17) What is the magneto motive force in a 75 turn coil of
wire when are 4 amps of current through it?**

**Given: –**

N = 75

I = 4

**Solutions: –**

MMF = N * I

= 75 * 4 = 300

**Answer: 300 At**

**18) If it takes 35 Jules of energy to move a charge of 6C
from one point to another. What is the voltage between the two points?**

**Given: –**

Q = 35

C = 6

**Solutions: –**

Q = C * V

V = Q / V = 35 / 6 = 5.83V

**Answer: V = 5.83V**

**19) Calculate the velocity of conductor if terminal voltage
is 2.2V/m & the magnetic field strength is 0.2T?**

**Given: –**

E = 2.2

L = 1

B = 0.2

**Solutions: –**

E = B * L * v

v = E / B*L = 2.2 / 0.2 = 11m/s

**Answer: v = 11 m/s**

**20) Calculate the frequency of the motor if the speed is
4800 rpm & no of poles are 2.**

**Given: –**

n = 4800 rpm

p = 2

**Solutions: –**

n = 120 f / p

f = 4800 * 2 / 120 = 80 Hz

**Answer: f = 80 Hz**

**21) What is the power in a mili watt having resistance of
12Ω having a potential difference of 3V?**

**Given: –**

R = 12Ω

V = 3V

**Solutions: –**

V = IR

I = V / R = 3 / 12 = ¼ A

P = V * I = 3 * ¼ = 0.75 Watts = 750 mW

**Answer: P = 750 mW**

**22) A 40 cm wire current of 20 amps is placed in a uniform
magnetic field of 30 degrees. Calculate the force on wire?**

**Given: –**

L = 40cm = 0.4 M

I = 20 A

B = 0.6

Ɵ = 30o

**Solutions: –**

F = B I L SinƟ

F = 0.6 * 20 * 0.4 * Sin 30o = 2.4N

**Answer: F = 2.4 N**

**23) Find the voltage of a motor of 200 rpm. When the flux
changes from 1wb to 4 wb?**

**Given: –**

N = 200

**Solutions: –**

E = N * (df/dt)

E = 200 * (4-1) = 600V

**Answer: E = 600 V**

**24) Inductance is 6mh current go from 0.4 to 2.5 amps in
0.6 sec. calculate induced emf?**

**Given: –**

L = 6 x 10-3 h

**Solutions: –**

E = L (di/dt)

E = 6 x 10-3 * (2.5 -0.4) / O.6 = 0.021V

**Answer: E = 0.021 V**

**25) The voltage induced across a certain coil is 150mV. A
100Ω resistor is connected to the coil terminal. Calculate the induced current
in mili amps?**

**Given: –**

R = 100Ω

V = 150mV = 0.150 V

**Solutions: –**

V = IR

I = V/R = 0.15/100 = 0.0015 A = 1.5 mA

**Answer: I = 1.5 mA**

**26) If the flux through 120 turns coil changes from 2wb to
6wb in one sec. what is the voltage induced?**

**Given: –**

N = 120

**Solutions: –**

E = N * (df/dt)

E = 120 * (6-2) = 480V

**Answer: E = 480 V**

**27) A 20m wire carrying a current of 10 amps magnetic field
of 0.3T with an angle of 40o with vector B. find the magnitude of force?**

**Given:** –

L = 20cm = 0.2 M

I = 10 A

B = 0.3

Ɵ = 40o

**Solutions: –**

F = B I L SinƟ

F = 0.3 * 10 * 0.2 * Sin 40o = 38.567 N

**Answer: F = 38.567 N**

**28) If the resistance of the wire is 15Ω, if the length is
doubled and dia is one third then resistance of wire is?**

R1 = 15 Ω

L2 = 2 L1

D2 = 1/3 D1

**Solutions: –**

R = ΡL/A

R2 = R1 * (A1 * L2) / (A2 * L1) = 15 * {( ) * 2 L1} / {[ ] * L1}

= 15 * 2 / (1/3)2 = 270 Ω

**Answer: R2 = 270 Ω**

**29) The resistors 40Ω, 50 Ω, 60 Ω are connected in series
input 30V. calculate voltage drop across 60Ω?**

**Given: –**

R1 = 40 Ω

R2 = 50 Ω

R3 = 60 Ω

V = 30V

**Solutions: –**

R = R1 + R2 + R3 = 40 + 50 + 60 = 150 Ω

V = IR

I = V / R = 30 / 150 = 0.2A

Voltage drop on R3 = I * R3

= 0.2 * 60 = 12 V

**Answer: Voltage drop on R3 = 12 V**

**30) A half wave rectifier having resistive load of 600 Ω
with supply voltage of 80V and having forward resistance of 100 Ω. Calculate
the peak current.**

**Given: –**

Vrms = 80V

RL = 600 Ω

RF = 100 Ω

**Solutions: –**

IM = VM / [RL + RF]

Vrms = VM / √2 => VM = Vrms * √2

IM = 80 X √2 / [600+100] = 0.1616 A

**Answer: IM = 0.1616 A**

**31) A certain appliances 700 W. if it is allowed to run
continuously for 36 days how many KW-hours of energy does it consumes?**

**Solutions: –**

Current consumes per hour = 700 W

For 15 days = 36 * 24 * 700 = 604800 W-h = 604.8KWh

**Answer: 604.8 KWh**

**32) Alfa particles enter an electromagnetic field where the
electric intensity is 300V/m and the magnetic induction is 0.200T. what is the
velocity of Alfa particles?**

**Given: –**

B = 0.200T

E = 300 V/M

**Solutions: –**

v = E / B = 300 / 0.2 =1500 m/s

**Answer: v = 1500 m/s**

**33) What is the magneto motive force in a 100 turns coil of
wire when there are 8A of current through it?**

**Given: –**

N = 100

I = 8

**Solutions: –**

MMF = N * I

= 100 * 8 = 800

**Answer: 800 At**

**34) A resistance of 20 Ω, 15 Ω, 25 Ω are connected in
series across a 36V line. Calculating voltage drop across 25 Ω resistance.**

**Given: –**

R1 = 20 Ω

R2 = 15 Ω

R3 = 25 Ω

V = 36V

**Solutions: –**

R = R1 + R2 + R3 = 20 + 15 + 25 = 60 Ω

V = IR

I = V / R = 36 / 600 = 0.6 A

Voltage drop on R3 = I * R3

= 0.6 * 25 = 15V

**Answer: Voltage drop on R3 = 15 V**

**35) The inductors of inductance 3H and 2H are connected in
series. The mutual inductance between them is 2H, which is aiding. Calculate
the effective inductance.**

**Given: –**

L1 = 3

L2 = 2

M = 2

**Solutions: –**

Inductors are connected in aeries which aiding

L = L1 + L2 + 2M

L = 3 + 2 + (2*2) = 9

**Answer: L = 9 H**

**36) The inductors of inductance 6H and 8H are connected in
series. The mutual inductance between them is 3H, which is aiding. Calculate
the effective inductance.**

**Given: –**

L1 = 6

L2 = 8

M = 3

**Solutions: –**

Inductors are connected in aeries which aiding

L = L1 + L2 + 2M

L = 6 + 8 + (2*3) = 20

**Answer: L = 20 H**

**37) Resistance of certain wire is 8Ω and second wire of the
same materials are same temp has a diameter half of the first wire and length
is 3 times the first wire. Find the resistance for second one.**

**Given: –**

R1 = 8 Ω

L2 = 3 L1

D2 = 0.5 D1

**Solutions: –**

R = ΡL/A

R2 = R1 * (A1 * L2) / (A2 * L1) = 8 * {( ) * 3 L1} / {( ) * L1}

= 8 * 3 / (0.5)2 = 96 Ω

**Answer: R2 = 96 Ω**

**38) 45Ω resistor connected to a 3V battery. Calculate the
power.**

**Given: –**

R = 45Ω

V = 3V

**Solutions: –**

V = IR

I = V/R = 3/45 = 1/15 A

P = V * I = 3 * 1/15 = 0.2W = 200 mW

**Answer: P = 200 m W**

**39) Alfa particles enter an electromagnetic field where the
electric intensity is 600V/m and the magnetic induction is 0.150T. what is the
velocity of Alfa particles?**

**Given: –**

B = 0.150T

E = 600 V/M

**Solutions: –**

v = E / B = 600 / 0.15 =4000 m/s

**Answer: v = 4000 m/s**

**40) 600 rps 2 pole find frequency?**

**Given: –**

n = 600 rps = 600 * 60 rpm

p = 2

**Solutions: –**

n = 120 f / p

f = 600 * 60 * 2 / 120 = 600 Hz

**Answer: f = 600 Hz**

**41) Electron having energy of 35J having a charge of 7C.
what is the voltage?**

**Given: –**

Q = 35

C = 7

**Solutions: –**

Q = C * V

V = Q / V = 35 / 7 = 5 V

**Answer: V = 5 V**

**42) The armature resistance of a 200V DC machine has 0.5Ω
of the full load armature current is 50A. what will be the induced emf when
machine act as**

**(i) Generator (ii) Motor**

**Given: –**

R = 0.5 Ω

V = 200 V

I = 50 A

**Solutions: –**

For Generator E = V + (I * R)

E = 200 + (50 * 0.5) = 225 V

For Motor E = V – (I * R)

E = 200 – (50 * 0.5) = 175 V

**Answer: (i) Generator E = 225 V (ii) Motor E = 175 V**

**43) electrons and protons of the same momentum enter
normally into a uniform magnetic field of flux density B. mass of the electron
is me (m suffix e) and mass of the proton is mp (m suffix p). if re (r suffix
e) and rp (r suffix p) represents the radii of the path of the electrons and
the protons respectively. calculate the ratio re/rp. your answer should be
correct to 2 decimals**

**Solutions: –**

Charged particles describe an anti-clockwise circular path and magnetic force work as centripetal force. Thus

F = qvB = mV2/R => mV/R

Here mpV and meV are same because of same momentum.

For electron qB = meV/Re

For proton qB = mpV/Rp

Dividing the both equations

Re / Rp = 1

**Answer: Re / Rp = 1.00**

**44) the certain instrument power is 4W and resistance is
100Ω. Calculate the voltage?**

**Given: –**

R = 100 Ω

P = 4W

**Solutions: –**

V = IR

I = V/R = V/100 A

P = V * I

4 = V * V/100 = V2/100

V2 = 400 => V = √400 = 20 V

**Answer: V = 20 V**

** 45) Find the current
flow through a light bulb from a steady movement of 1022 electrons in one hour.
Assume that the change of a single electron = 1.602×10-19**

**Given: –**

n = 1022

e = 1.602×10-19

t = one hour = 60 * 60 Sec

**Solutions: –**

q = n * e = 1022 * 1.602×10-19

I = q / t = (1022 * 1.602×10-19 ) / (60 * 60) = 0.445 A

**Answer: I = 0.445 A**

**46) The types of the system whose transfer function is
given by (0 Decimal)**

G(S) = (S2+3) / (S5+S4+S3+3S2+2S) is: –

**Answer: 1**

**47) If V = 3v, R= 120ohms. Then find I in mili amp?**

**Given: –**

R = 120Ω

V = 3V

**Solutions: –**

V = IR

I = V/R = 3/120 = 0.025 A = 25 mA

**Answer: I = 25 mA**

**48) Find the current through a light bulb from a steady
movement of 1022 electrons in 0.25 hours. Assume the charge of single electron
= 1.602×10^-19 coulombs (correct to 3 decimal)**

**Given: –**

n = 1022

e = 1.602×10-19

t = 0.25 hour = 0.25 * 60 * 60 Sec

**Solutions: –**

q = n * e = 1022 * 1.602×10-19

I = q / t = (1022 * 1.602×10-19 ) / (0.25 * 60 * 60) = 1.78 A

**Answer: I = 1.780 A**

**49) The total secondary voltage in a center tapped full
wave rectifier is 125 Vrms. neglecting the diode drop the rms output voltage
is?**

**Solutions: –**

Output voltage without diode drop = Vrms / 2 = 125/2 = 62.5 V

**Answer: Vdc = 62.5 V**