1. ELE101/102 Dept of E&E,MIT Manipal 1
Tutorial
1. A resistance of 50Ω 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:
Ω°∠ 14.3259
A14.32898.3 °−∠
0.847 lag
759.15W
IVR
VL
V
32.12º

2. ELE101/102 Dept of E&E,MIT Manipal 2
Tutorial
2. A resistance of 50Ω is connected in series with a
capacitance of 100 µF 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:
Ω°−∠ 48.32272.59
A°∠ 48.3288.3
0.843 lead
752.81 W
I
VR
VC
V
32.48º

3. ELE101/102 Dept of E&E,MIT Manipal 3
Tutorial
3. The value of the capacitor in the circuit given below is
20 µF 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.
C
R L RLI
V
25 V
40 V
55 V
coil
50 V
degree26.768byVleadsIcurrent34.155V;V50Hz,f
:Ans
==
Power Loss = 1.8967 W

4. 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 Ω.
R = Z cos φ; R = 8.15 Ω
XC = 14.11 Ω; C= 169.6 microfarad

5. 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
L
L
coil
125 V 200 V
250 V, 50 Hz

6. ELE101/102 Dept of E&E,MIT Manipal 6
Tutorial
Phasor Diagram
IVR = IR IRL
IXL
V
Vcoil
R = 25 Ω
Zcoil = 40 Ω
RL = 5.5 Ω
XL = 39.62 Ω
Total pf = 0.61
Total power = 762.5 W
Power absorbed by the coil = 137.5 W

7. 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 H
6 Ω
R C
I
Vb
Va
V=240V, 50 Hz
I - ref
Vb
Va
θa
θb=53.16º
Zb = 10∠53.16º
Since Vb and Va are in quadrature.
Za = 3.336∠-90+53.16º
= 3.336∠-36.84º
R = 2.669 Ω
FC 3
1059.1 −
×=
Za = Zb / 3 = 3.336
Solution