1. ELE101/102 Dept of E&E,MIT Manipal 1
1. An impedance of (2+j6) Ω is connected in series with
two impedances (10+j4) Ω and (12-j8) Ω which are in
parallel. When this combination Is connected across
250V AC supply, calculate a) current drawn from the
supply. b) pf of the circuit. c) total power absorbed. d)
power dissipated in each resistance. Draw the phasor
diagram.
10+j4
12-j8
2+j6
I
ZA
Z1
Z2
V = 250V
I1
I2
V1 V2
Tutorial

2. ELE101/102 Dept of E&E,MIT Manipal 2
Zeq = ZA + Z1//Z2
Zeq = 10.66∠33º Ω.
I = 23.45∠-33º A
Pf = 0.84 lag
P = 4916.7 W
I1 = 15.12∠-56.4º A I2 = 11.3∠-0.9º A
P2 Ω = 1099.8 W
P10 Ω = 2286.1 W
P12Ω = 1532.3 W
Tutorial

3. ELE101/102 Dept of E&E,MIT Manipal 3
Phasor Diagram
I1
56.4º
I2
0.9º
33º
I
VRA
VLA
V1
V2
V
Tutorial

4. ELE101/102 Dept of E&E,MIT Manipal 4
2. Two circuits having the same numerical impedances
are connected in parallel. The power factor of one
circuit is 0.8 lag and of the other is 0.6 lag. Determine
the power factor of the parallel combination.
Ans = 0.707 lag
Tutorial

5. ELE101/102 Dept of E&E,MIT Manipal 5
3. An inductive circuit supplied with 250V, 50 Hz has an
active power of 11.9 KW and apparent power of 17
KVA. a) Find the power factor of the circuit. b) Draw
the power triangle. c) Find the value of the
capacitance required to improve the pf to 1, 0.9 lag
and 0.9 lead.
Illustration
45.57º
11.9 KW
12.14 KVAR
17 KVA
a) pf = 0.7 lag
b) Power Triangle
Ans:

6. ELE101/102 Dept of E&E,MIT Manipal 6
VIC sinφC =
3
1014.12 ×
IC= 48.56∠90º A
XC = 5.148∠-90º Ω
C = 618.3 µF
For cos φ = 0.9 lag:
25.84º
11.9 KW
Xp
17 KVA
Xp = 5.76 KVAR
VIC sinφC = 3
1038.6 ×
C = 324.9 µF
For cos φ = 0.9 lead: C = 911.6 µF
For cos φ = 1:

7. ELE101/102 Dept of E&E,MIT Manipal 7
Illustration
(4) Inductive loads of 0.8KW and 1.2KW at pf of 0.8 and 0.6
respectively are connected in parallel across a 200V, 50Hz
supply. Calculate the total current taken by the
combination. Draw the power triangle & find the value of
the capacitance to be connected in parallel to improve the
power factor to a) 0.9 lag b) 0.9 lead
Ans:
IT=14.866A
a) 0.9 lag C = 98 µF
b) 0.9 lead C = 252 µF