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 t --------------- (1)

= Im sint--------------(2)

i(t)

v(t) R

VIPhasor diagram

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

Pure Resistive Circuits -Power Consumed

Instantaneous power,

p(t) = v(t).i(t)

= Vm Im sin2t

T

0

dt)t(pT

1P,PowerAverage

2t

0 /2

p(t)

v(t)

i(t)

VI2

IV mm

R

VRIP

22

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

Pure Inductive circuits

i(t)

v(t) L

+

-

e(t)

00

mm

m

m

m

m

90I ;0V

;ωL

VI where

(2)........eq 90)-t(sinI i(t)

)c(ωL

V

dt t sinVL

1 dt v(t)

L

1 i(t)

(1)........eqt sinVv(t)

Let

IV

tos

t/2 3/2 2

v(t)

i(t)

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

Pure Inductive circuits…Phasor diagramCurrent through pure inductor lags the voltage across it by 90º.

Vm = L Im.

= XL Im.

XL = L = 2f L ----- inductive reactance, in ohm

Phasor Diagram

L

mm

L

mm X

2/V

2

I

X

VI

I

VX

X

VI L

L

90º

0VV

90II

LL XI

VjX

I

V

I

V

where90

00

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

Pure Inductive Circuits -Power Consumed

Instantaneous power,

p(t) = v(t).i(t)

= Vm Im sint. sin(t-90)

= - Vm Im cost. sint

= -(Vm Im/2) sin2t

Average Power Consumed in a Pure Inductor

t/2 3/2 2

v

i p

0

02

1 2

0

)t(d)t(pP

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

Pure Capacitive circuits

00

m

m

m

m

90I ;0V

;I where

.....eq(2) 90)t(sinI i(t)

)(c dt

t)sind(VC i(t)

;dt

dv(t)i(t)

.....eq(1)t sinVv(t)

Let

IV

CV

tosCV

C

m

m

i(t)

v(t)C

v(t)

i(t)

t/2 3/2 2

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/(C)

= XC Im.

XC = 1/(C) = 1/(2f C) ----- Capacitive reactance, in ohm

Phasor Diagram

C

mm

C

mm X

/VI

X

VI

2

2

I

VX

X

VI C

C

90º 0VV

90II

CC XI

VjX

I

V

I

V

where90

00

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

Pure Capacitive Circuits -Power Consumed

Instantaneous power,

p(t) = v(t).i(t)

= Vm Im sint. sin(t+90)

= Vm Im cost. sint

= (Vm Im/2) sin2t

Average Power Consumed in a Pure Capacitor

02

1 2

0

)t(d)t(pP

t/2 3/2 2

v

i p

0