In this PowerPoint Presentation we study about the analysis of an AC circuit: RC circuit i.e. a circuit containing a resistor and indicator in series connected to an AC input. Follow for more presentations.

1. PowerPoint Presentation by:- Mohit

2. What is an RC circuit ?
 A circuit which contains a pure resistance R ohms
connected in series with a pure capacitor of
capacitance C farads is known as RC circuit.

3. Steps to Draw Phasor Diagram
1. Current I(r.m.s value) is taken as
reference.
2. Voltage drop in resistance is Vr. (Vr=IR
is drawn in phase with the current I.)
3. Voltage drop in capacitive reactance is
Vc. (Vc = Ixc and is drown 90 degree
behind the current.)
4. The vector sum of the two voltage drop
is equal to the applied voltage (V).

4.  Vr = IR and Vc = IXc (where Xc = 1/2πfC
 In right angled triangle OAB
V = √(Vr)2 + (Vc)2 = √(IR)2 + (IXc)2 = I√R2 + Xc2
Or I = V/√R2 +Xc2 = V/Z ………..( 1 )
Where Z = √R2 + Xc2 is the total opposition offered to
the flow of alternating current by an RC circuit and is
called impedence of the circuit.
Current in RC Circuit

5.  From the phasor diagram it is clear that current in this
circuit leads the applied voltage by an angle ø called
phase angle.
 From the phasor diagram
tan Ф = Vc/Vr = IXc/IR = Xc/R
Or Ф = tan-1 Xc/R

6.  If the alternating voltage applied across the circuit is given by the
equation:
v = Vm Sin ωt
Then , i = Im Sin(ωt + Ф)
Instantaneous power,
P = vi = Vm Sin ωt . Im Sin(ωt + Ф)
= (VmIm/2) 2 Sin(ωt + Ф) sin ωt
= (Vm/√2)(Im/√2)[ Cos Ф – Cos(2 ωt + Ф)]
= (Vm/√2)(Im/√2) Cos Ф - (Vm/√2)(Im/√2) Cos(2 ωt + Ф)

7. Average Power
 Average power consumed in the circuit over a complete
cylce,
P = average of (Vm/√2)(Im/√2) Cos Ф –
average of (Vm/√2)(Im/√2) Cos(2 ωt + Ф)
Or P = (Vm/√2)(Im/√2) Cos Ф – zero `
Or P = Vr.m.s . Ir.m.s Cos Ф = VI Cos Ф
where Cos Ф is called power factor of the circuit.

8.  From Phasor diagram
Cos Ф = Vr/V = IR/IZ = R/Z ………..( 2 )
Putting the value of V from eqn 1 and Cos Ф from eqn
the value of power will be
P = (IZ)(I)(R/Z) = I2R ………..(3)
From eqn 3 it is observed that power is actually
consumed in resistance only , capacitor does not
consume any power.

10.  The various points on the power curve are obtained
from the product of the instantaneous value of voltage
and current.
 The power is negative between the angle (180o - Ф) and
180o and between (360o – Ф) and 360o .
 In rest of the cycle, the power is positive.
 Area under the positive loops > negative loops,
 the net power over a complete cycle is positive.
 A definite quantity of power is utilisesd or consumed
by this circuit.

11.  When each side of the simplified phasor diagram is
divided by a common factor I, we get another right
angled triangle known as Impedence triangle.
Phasor Diagram Impedence Triangle