![ 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)](https://image.slidesharecdn.com/presentation1-200605163603/85/Presentation-of-Electrical-Engineering-Analysis-of-AC-circuit-RC-circuit-6-320.jpg)
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Presentation of Electrical Engineering ( Analysis of AC circuit: RC circuit )
- 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