ELE 102/102 Dept of E&E MIT Manipal 1

Tutorial

1) Find the Average value and RMS value of the given non sinusoidal waveform shown below.

0

A

-A

T/4 T/2 T t

Ans: Iav= A / 2

3

AIRMS

ELE101/102 Dept of E&E,MIT Manipal 2

Tutorial

2. A resistance of 50 is connected in series with an inductance of 100 mH

across a 230V, 50 Hz, single phase AC supply. Calculate

a) Impedance b) current drawn c) power factor d) power consumed

e) Draw the phasor diagram.

Ans:

14.3259

A14.32898.3

0.847 lag

759.15WIVR

VLV

32.12º

ELE101/102 Dept of E&E,MIT Manipal 3

Tutorial

3. A resistance of 50 is connected in series with a capacitance of 100

F across a 230V, 50 Hz, single phase AC supply. Calculate

a) Impedance b) current drawn c) power factor d) power consumed

e) Draw the phasor diagram.

Ans:

48.32272.59

A 48.3288.3

0.843 lead

752.81 W

I

VR

VC

V

32.48º

ELE101/102 Dept of E&E,MIT Manipal 4

Tutorial

4. The value of the capacitor in the circuit given below is 20 F and the

current flowing through the circuit is 0.345 A. If the voltages are as

indicated, find the applied voltage, the frequency and loss in the

coil.

CR L RLI

V25 V

40 V55 V

coil

50 V

degree 26.768by V leads Icurrent 34.155V; V

50Hzf

:Ans

Power Loss = 1.8967 W

ELE101/102 Dept of E&E,MIT Manipal 5

Tutorial

5. An emf of v= 326 sin 418t is applied to a circuit. The

current is i = 20 sin(418t + 60). Find the circuit

components, frequency of the input voltage and

power factor.

Solution:

f=66.5 Hz

pf = 0.5.

Z = Vm / Im = 16.3 .

R = Z cos ; R = 8.15 XC = 14.11 ; C= 169.6 microfarad

ELE101/102 Dept of E&E,MIT Manipal 6

Tutorial

6. A current of 5 A flows through a non inductive

resistance in series with a coil when supplied at 250

V, 50 Hz. If the voltage across the resistance is 125

V, calculate

a) the impedance, reactance and resistance of the coil.

b) power absorbed by the coil.

c) Total power. Draw the phasor diagram.

R RL LL

coil

125 V 200 V

250 V, 50 Hz

ELE101/102 Dept of E&E,MIT Manipal 7

Tutorial

Phasor Diagram

R RL LL

coil

125 V 200 V

250 V, 50 Hz

IVR = IR IRL

IXL V

Vco

il

R = 25 Zcoil = 40

RL = 5.5

XL = 39.62

Total pf = )/(

)(

IV

RR L= 0.61

Total power = 762.5 W

Power absorbed by the coil = 137.5 W

ELE101/102 Dept of E&E,MIT Manipal 8

Tutorial

6. Find the values of R and C so that Vb = 3 Va and Vb and Va are in quadrature. Find also the phase relation between V and Va , Va and I.

0.0255 H6 R C

IVb

Va

V=240V, 50 Hz

I - ref

Vb

Va

a

b=53.16º

Zb = 1053.16º

Since Vb and Va are in quadrature.

Za = 3.336-90+53.16º = 3.336-36.84º

R = 2.669

FC 31059.1

Za = Zb / 3 = 3.336

Solution