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Ac current

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Alternating current (AC), is an electric current in which the flow of electric charge periodically reverses direction, whereas in direct current (DC, also dc), the flow of electric charge is only in one direction.

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Alternating Current Resistance, Inductance and Capacitance
AC through resistance, inductance and capacitance • AC through ohmic resistance alone Let the voltage v(t) = Vm sin ωt is applied to a pure resistor R Let i = instantaneous current. Applied voltage has to supply ohmic voltage drop only. Hence, for equilibrium, v = iR = Vm sin ωt •Current, i is maximum when sin ωt is unity. So • Hence i = Im sin ωt • Alternating voltage and current are in phase with each other
AC through resistance, inductance and capacitance • Power in ohmic resistance alone Instantaneous power, p = v i = (Vm sin ωt)(Im sin ωt) = Vm Im sin2 ωt = (Vm Im/2) (1- cos 2ωt) = Vm Im/2 - (Vm Im/2) cos 2ωt • Power consists of a constant part = Vm Im/2 and • a fluctuating part (Vm Im/2) cos 2ωt of double frequency For a complete cycle, the average value of (Vm Im/2) cos 2ωt = 0 Hence, power for the whole cycle is P = (Vm Im/2) = (Vm/√2) Im/√2) P = V x I watt Where V = rms value of applied voltage and I = rms value of the current
AC through resistance, inductance and capacitance • AC through ohmic resistance alone
AC through resistance, inductance and capacitance • AC through ohmic resistance alone Ex: A 60 Hz voltage of 115 V (r m s) is impressed on a 10 Ω resistor. (i) Write down the time expression (equation) of voltage v(t) and the resulting current i(t). Sketch v(t) and i(t).
AC through resistance, inductance and capacitance AC through pure inductance alone • Whenever an alternating voltage is applied to a purely inductive coil, a back emf is produced due to the self inductance of the coil. • The back emf at every step, opposes the rise or fall of current through the coil. • The applied voltage has to overcome this self induced emf only as there is no ohmic voltage drop, So at every step
AC through resistance, inductance and capacitance AC through pure inductance alone Now So Hence,
AC through resistance, inductance and capacitance AC through pure inductance alone If applied voltage Then resulting current Vectorially • Clearly, the current lags behind the applied voltage by a quarter cycle (as shown, or the phase difference between is π/2 with voltage leading. • Here, Im = (Vm/ωL) = (Vm/XL) . Hence ωL plays the part of resistance. It is called the inductive reactance XL of the coil and is given in ohms if L is in henry and ω is in radian/second.
AC through resistance, inductance and capacitance AC through pure inductance alone •current lags behind the applied voltage by π/2 • inductive reactance XL= ωL Power • It is clear that the average demand of power from the supply for a complete cycle is zero. Power wave is a sine wave of double frequency. Maximum instantaneous power is
AC through resistance, inductance and capacitance AC through pure inductance alone •current lags behind the applied voltage by π/2 • inductive reactance XL= ωL • average demand of power from the supply is zero • Energy is stored in one half cycle and • releases in the next half cycle
AC through pure inductance Ex: A sinusoidal voltage of v = 100 Sin 20t is Applied across a 0.5 H inductor. What is the sinusoidal expression for the current? Inductive reactance, Ex: A sinusoidal voltage of v = 100 Sin 20t is Applied across a 0.5 H inductor. What is the sinusoidal expression for the current? Inductive reactance,
AC through pure inductance
AC through resistance, inductance and capacitance AC through pure capacitance Whenever an alternating voltage is applied to the plates of a capacitor, the capacitor is charged first in one direction and then in the opposite direction We know,
AC through resistance, inductance and capacitance AC through pure capacitance
AC through resistance, inductance and capacitance AC through pure capacitance In Ohm’s law format • Capacitive reactance is the opposition to the flow of charge, which results in the continual interchange of energy between the source and the electric field of the capacitor. • Like the inductor, the capacitor does not dissipate energy in any form
AC through resistance, inductance and capacitance AC through pure capacitance
AC through resistance, inductance and capacitance AC through pure capacitance
AC through resistance, inductance and capacitance AC through pure capacitance Ex: A voltage of v = 3- Sin 400t is applied across a 1 µF capacitor. What is the sinusoidal expression for the current?
AC through resistance, inductance and capacitance AC through pure capacitance Ex: A current I = 40 Sin (500t +60 ) is impressed on a 100 µF capacitor. What is the sinusoidal expression for voltage?
AC through resistance, inductance and capacitance AC through pure capacitance: Power Instantaneous power p = vC iC = Vm sin ωt.Im sin(ωt + π/2). = Vm Im sinωt cos ωt = ½ Vm Im sin 2ωt Power for the whole cycle = ½ Vm Im sin 2ωt dt = 0 So, in a purely capacitive circuit, the average power demand from power supply is zero. Also power wave is a sine wave of frequency double that of the voltage and current waves. Maximum value of the instantaneous power is = Vm Im/2.

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Ac current

  • 2. AC through resistance, inductance and capacitance • AC through ohmic resistance alone Let the voltage v(t) = Vm sin ωt is applied to a pure resistor R Let i = instantaneous current. Applied voltage has to supply ohmic voltage drop only. Hence, for equilibrium, v = iR = Vm sin ωt •Current, i is maximum when sin ωt is unity. So • Hence i = Im sin ωt • Alternating voltage and current are in phase with each other
  • 3. AC through resistance, inductance and capacitance • Power in ohmic resistance alone Instantaneous power, p = v i = (Vm sin ωt)(Im sin ωt) = Vm Im sin2 ωt = (Vm Im/2) (1- cos 2ωt) = Vm Im/2 - (Vm Im/2) cos 2ωt • Power consists of a constant part = Vm Im/2 and • a fluctuating part (Vm Im/2) cos 2ωt of double frequency For a complete cycle, the average value of (Vm Im/2) cos 2ωt = 0 Hence, power for the whole cycle is P = (Vm Im/2) = (Vm/√2) Im/√2) P = V x I watt Where V = rms value of applied voltage and I = rms value of the current
  • 4. AC through resistance, inductance and capacitance • AC through ohmic resistance alone
  • 5. AC through resistance, inductance and capacitance • AC through ohmic resistance alone Ex: A 60 Hz voltage of 115 V (r m s) is impressed on a 10 Ω resistor. (i) Write down the time expression (equation) of voltage v(t) and the resulting current i(t). Sketch v(t) and i(t).
  • 6. AC through resistance, inductance and capacitance AC through pure inductance alone • Whenever an alternating voltage is applied to a purely inductive coil, a back emf is produced due to the self inductance of the coil. • The back emf at every step, opposes the rise or fall of current through the coil. • The applied voltage has to overcome this self induced emf only as there is no ohmic voltage drop, So at every step
  • 7. AC through resistance, inductance and capacitance AC through pure inductance alone Now So Hence,
  • 8. AC through resistance, inductance and capacitance AC through pure inductance alone If applied voltage Then resulting current Vectorially • Clearly, the current lags behind the applied voltage by a quarter cycle (as shown, or the phase difference between is π/2 with voltage leading. • Here, Im = (Vm/ωL) = (Vm/XL) . Hence ωL plays the part of resistance. It is called the inductive reactance XL of the coil and is given in ohms if L is in henry and ω is in radian/second.
  • 9. AC through resistance, inductance and capacitance AC through pure inductance alone •current lags behind the applied voltage by π/2 • inductive reactance XL= ωL Power • It is clear that the average demand of power from the supply for a complete cycle is zero. Power wave is a sine wave of double frequency. Maximum instantaneous power is
  • 10. AC through resistance, inductance and capacitance AC through pure inductance alone •current lags behind the applied voltage by π/2 • inductive reactance XL= ωL • average demand of power from the supply is zero • Energy is stored in one half cycle and • releases in the next half cycle
  • 11. AC through pure inductance Ex: A sinusoidal voltage of v = 100 Sin 20t is Applied across a 0.5 H inductor. What is the sinusoidal expression for the current? Inductive reactance, Ex: A sinusoidal voltage of v = 100 Sin 20t is Applied across a 0.5 H inductor. What is the sinusoidal expression for the current? Inductive reactance,
  • 12. AC through pure inductance
  • 13. AC through resistance, inductance and capacitance AC through pure capacitance Whenever an alternating voltage is applied to the plates of a capacitor, the capacitor is charged first in one direction and then in the opposite direction We know,
  • 14. AC through resistance, inductance and capacitance AC through pure capacitance
  • 15. AC through resistance, inductance and capacitance AC through pure capacitance In Ohm’s law format • Capacitive reactance is the opposition to the flow of charge, which results in the continual interchange of energy between the source and the electric field of the capacitor. • Like the inductor, the capacitor does not dissipate energy in any form
  • 16. AC through resistance, inductance and capacitance AC through pure capacitance
  • 17. AC through resistance, inductance and capacitance AC through pure capacitance
  • 18. AC through resistance, inductance and capacitance AC through pure capacitance Ex: A voltage of v = 3- Sin 400t is applied across a 1 µF capacitor. What is the sinusoidal expression for the current?
  • 19. AC through resistance, inductance and capacitance AC through pure capacitance Ex: A current I = 40 Sin (500t +60 ) is impressed on a 100 µF capacitor. What is the sinusoidal expression for voltage?
  • 20. AC through resistance, inductance and capacitance AC through pure capacitance: Power Instantaneous power p = vC iC = Vm sin ωt.Im sin(ωt + π/2). = Vm Im sinωt cos ωt = ½ Vm Im sin 2ωt Power for the whole cycle = ½ Vm Im sin 2ωt dt = 0 So, in a purely capacitive circuit, the average power demand from power supply is zero. Also power wave is a sine wave of frequency double that of the voltage and current waves. Maximum value of the instantaneous power is = Vm Im/2.