1. ELE101/102 Dept of E&E,MIT Manipal 1
Pure resistive circuits
Phase angle between V & I = 0
current through the resistor is in phase with the voltage across it.
R
)t(v
)t(i =
Let
v(t) = Vm sin ωt --------------- (1)
= Im sinωt--------------(2)
i(t)
v(t) R
VIPhasor diagram

2. ELE101/102 Dept of E&E,MIT Manipal 2
Pure Resistive Circuits -Power Consumed
Instantaneous power,
p(t) = v(t).i(t)
= Vm Im sin2
ωt
∫=
T
0
dt)t(p
T
1
P,PowerAverage
2π
ωt
0 π/2 π
p(t)
v(t)
i(t)
VI
2
IV mm
==
R
V
RIP
2
2
==

3. ELE101/102 Dept of E&E,MIT Manipal 3
Pure Inductive circuits
i(t)
v(t) L
+
-
e(t)
00
m
m
m
m
m
m
90I;0V
;
ωL
V
Iwhere
(2)........eq90)-t(sinIi(t)
)c(
ωL
V
dttsinV
L
1
dtv(t)
L
1
i(t)
(1)........eqtsinVv(t)
Let
−∠=∠=
=
=
−=
==
=
∫∫
IV
tos
ω
ω
ω
ω
ωtπ/2 π 3π/2 2π
v(t)
i(t)

4. ELE101/102 Dept of E&E,MIT Manipal 4
Pure Inductive circuits…Phasor diagram
Current through pure inductor lags the voltage across it by 90º.
Vm = ωL Im.
= XL Im.
XL = ωL = 2πf L ----- inductive reactance, in ohm
Phasor Diagram
L
mm
L
m
m
X
2/V
2
I
X
V
I ==
I
V
X
X
V
I L
L
==
90º
°∠= 0VVˆ
°−∠= 90IIˆ
LL X
I
V
jX
I
V
I
V
==
−∠
∠
= where
90
0
0

5. ELE101/102 Dept of E&E,MIT Manipal 5
Pure Inductive Circuits -Power Consumed
Instantaneous power,
p(t) = v(t).i(t)
= Vm Im sinωt. sin(ωt-90)
= - Vm Im cosωt. sinωt
= -(Vm Im/2) sin2ωt
Average Power Consumed in a Pure Inductor
ωt
π/2 π 3π/2 2π
v
i p
0
0
2
1
2
0
== ∫
π
ω
π
)t(d)t(pP

6. ELE101/102 Dept of E&E,MIT Manipal 6
Pure Capacitive circuits
00
m
m
m
m
90I;0V
;Iwhere
.....eq(2)90)t(sinIi(t)
)(c
dt
t)sind(V
Ci(t)
;
dt
dv(t)
i(t)
.....eq(1)tsinVv(t)
Let
∠=∠=
=
+=
=
=
=
=
IV
CV
tosCV
C
m
m
ω
ω
ωω
ω
ω
i(t)
v(t)
C
v(t)
i(t)
ωtπ/2 π 3π/2 2π

7. ELE101/102 Dept of E&E,MIT Manipal 7
Pure Capacitive circuits-Phasor diagram
Current through pure Capacitor leads the voltage across it by 90º.
Vm = Im/(ωC)
= XC Im.
XC = 1/(ωC) = 1/(2πf C) ----- Capacitive reactance, in ohm
Phasor Diagram
C
mm
C
m
m
X
/VI
X
V
I
2
2
==
I
V
X
X
V
I C
C
==
90º °∠= 0VVˆ
°∠= 90IIˆ
CC X
I
V
jX
I
V
I
V
=−=
∠
∠
= where
90
0
0

8. ELE101/102 Dept of E&E,MIT Manipal 8
Pure Capacitive Circuits -Power Consumed
Instantaneous power,
p(t) = v(t).i(t)
= Vm Im sinωt. sin(ωt+90)
= Vm Im cosωt. sinωt
= (Vm Im/2) sin2ωt
Average Power Consumed in a Pure Capacitor
0
2
1
2
0
== ∫
π
ω
π
)t(d)t(pP
ωt
π/2 π 3π/2 2π
v
i p
0