FUNCTION -5 (ELECTRICAL)

FAQ NUMERICALS

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

Solution: –

n = 120 f / p

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

Answer: f = 400 Hz

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

Given: –

VM = 12V

Solution: –

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 20ºC 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 10ºC

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

Solutions: –

At 30ºC IR is 10/2 = 5 MΩ

At 40ºC IR is 5/2 = 2.5 MΩ

Answer: 2.5 MΩ

7) Convert 0.06 KV into ______ mV

Solution: –

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

Solution: –

[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?

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

n = 120 f / p

f = 4800 * 2 / 120 = 80 Hz

Answer: f = 80 Hz

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

Given: –

R = 12Ω

V = 3V

Solution: –

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

Solution: –

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

Solution: –

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 miliamps?

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