ELE101/102 Dept of E&E,MIT Manipal 1

Tutorial

1. 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 2

Tutorial

2. 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:

48.32272.59A 48.3288.3

0.843 lead752.81 W

I

VR

VC

V

32.48º

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

Tutorial

3. 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 50Hz,f

:Ans

Power Loss = 1.8967 W

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

Tutorial

4. 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 5

Tutorial

5. 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 and that across the

coil is 200V, 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 6

Tutorial

Phasor Diagram

IVR = IR IRL

IXL V

Vco

il

R = 25

Zcoil = 40

RL = 5.5 XL = 39.62

Total pf = 0.61

Total power = 762.5 W

Power absorbed by the coil = 137.5 W

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

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