# MEO CLASS 4 – ELECTRICAL PROBLEMS

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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]

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]

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

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

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

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

7) Convert 0.06 KV into ______ mV

Solutions: –

0.06 * 1000 = 60V

60 * 1000 = 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

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

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

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: –

= (61.5 – 60) / 61.5 = 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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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