MMD PAPERS

# FUNCTION -5 (ELECTRICAL)

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

Solution: –

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

Solution: –

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?

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

Solution: –

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

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

Solution: –

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

Solution: –

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

Solution: –

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: – 1

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

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

Solution: –

q = n * e = 1022 * 1.602×10-19

I = q / t = (1022 * 1.602×10-19) / (0.25 * 60 * 60) = 1.78 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 • Akshay says: