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- 1. Welcome to Classes BYJUβS Alternating Current What you already know What you will learn S5: AC Circuits (Combination) 1 . Sinu so idal AC param e te rs 2 . Phasor diagram s 3 . Pu re re sistiv e AC circu its 4. Pu re capacitiv e AC circu its 5 . Pu re indu ctiv e AC circu its 1 . Im pe dance 2 . RC co m b inatio n 3 . L R co m b inatio n
- 2. Pure inductive AC circuit π = π0 sin ππ‘ β π 2 π0 = π0 πΏπ = π0 ππΏ SI Unit : Ohm (Ξ©) π = π0 sin ππ‘ πΏ ππΏ ππΏ = πΏπ Inductive reactance Phase difference π = ππ‘ β π 2 β ππ‘ = β π 2 Current lags potential by 90Β°
- 3. Pure inductive AC circuit π = π0 sin ππ‘ β π 2 π = π0 sin ππ‘ Phasor diagram π0 π¦ π₯ π0 ππ‘ Wave diagram π‘ π, π 0 π 2π π 2 3π 2 π = π0 sin ππ‘ π = π0 sin ππ‘ β π 2 π0 π0 π0
- 4. Pure capacitive AC circuit π = π0 sin ππ‘ πΆ π = π0 sin ππ‘ + π 2 π0 = π0 1 πΆπ = π0 ππΆ SI Unit : Ohm Ξ© ππΆ = 1 πΆπ Capacitive reactance Phase difference π = ππ‘ + π 2 β ππ‘ = π 2 Current leads potential by 90Β°
- 5. Pure capacitive AC circuit Phasor diagram π0 π¦ π₯ π0 ππ‘ Wave diagram π = π0 sin ππ‘ + π 2 π‘ π, π 0 π 2π π 2 3π 2 π = π0 sin ππ‘ π0 π0 π0
- 6. Simple AC circuits Element Current (emf:π = π0 sin ππ‘) π π(Reactance) Resistor Inductor Capacitor π0 π sin ππ‘ π0 ππΏ sin ππ‘ β π 2 π0 ΰ΅ 1 ππΆ sin ππ‘ + π 2 0 β π 2 π 2 π ππΏ 1 ππΆ
- 7. Wave diagram π‘ π, π 0 π 2π π 2 3π 2 π = π0 sin ππ‘ π = π0 sin ππ‘ β π 2 π0 π0 π0 0 β π 2 π 2 β π π β 3π 2 3π 2 β 2π π = π0 sin ππ‘ ππΏ π increases from zero to maximum, current becomes zero. π decreases from maximum to zero, current grows to maximum. π increases from zero to negative maximum, current becomes zero. π decreases from negative maximum to zero, current grows to negative maximum. Current tries to follow path of voltage. The phase difference between them always remains to be 90Β°.
- 8. π‘ π, π 0 π 2π π 2 3π 2 π = π0 sin ππ‘ π = π0 sin ππ‘ β π 2 π0 π0 π0 π‘ π, π 0 π 2π π 2 3π 2 π = π0 sin ππ‘ π = π0 sin ππ‘ β π 2 π0 π0 π0 The lagging of current can be seen from the mechanical analogy as shown. Height represents the voltage and velocity of block is current. When height is maximum, velocity is zero and vice-versa.
- 9. Wave diagram π‘ π, π 0 π 2π π 2 3π 2 π = π0 sin ππ‘ π = π0 sin ππ‘ β π 2 π0 π0 π0
- 10. π = π0 sin ππ‘ πΆ Wave diagram π = π0 sin ππ‘ + π 2 π‘ π, π 0 π 2π π 2 3π 2 π = π0 sin ππ‘ π0 π0 π0
- 11. π The relation b/w peak current and peak voltages can be written as Purely resistive circuit Purely inductive circuit π0 = E0 π π = π π = ππΏ π is called impedance. Impedance is defined as the opposition any circuit shows when voltage is applied to it. SI unit is Ohm (Ξ©) Purely capacitive circuit π = 1/ππΆ
- 12. β | E0 sin ππ‘ πΆ π E0 2 = ππ 2 + ππΆ 2 E0 2 = π0 2 π 2 + π0 2 ππΆ 2 π0 2 π2 = π0 2 π 2 + π0 2 ππΆ 2 π2 = π 2 + ππΆ 2 π = π 2 + ππΆ 2 π = π 2 + 1/ππΆ 2 ππΆ = 1/ππΆ β ππ π ππΆ E0 π0
- 13. β | π = π 2 + 1/ππΆ 2 tan π = ππΆ π π ππΆ π π E0 sin ππ‘ πΆ π tan π = 1 ππΆπ π2 = π 2 + ππΆ 2 ππΆ = 1 ππΆ Peak current (π0) is given by- π0 = E0 π = E0 π 2 + 1/ππΆ 2 ππ π ππΆ E0 π0
- 14. β | π = π 2 + 1/ππΆ 2 tan π = 1 ππΆπ π0 = E0 π = E0 π 2 + 1/ππΆ 2 Steady state current (π) in the circuit π = E0 π sin(ππ‘ + π) E0 sin ππ‘ πΆ π The current leads the emf by π. ππ π ππΆ E0 π0
- 15. β | E0 2 = ππ 2 + ππΏ 2 E0 2 = π0 2 π 2 + π0 2 ππΏ 2 π0 2 π2 = π0 2 π 2 + π0 2 ππΏ 2 π2 = π 2 + ππΏ 2 π = π 2 + ππΏ 2 π = π 2 + ππΏ 2 ππΏ = ππΏ β E0 sin ππ‘ πΏ π ππ π ππΏ E0 π0
- 16. β | tan π = ππΏ π π ππΏ π π tan π = ππΏ π ππΏ = ππΏ Peak current (π0) is given by- π0 = E0 π = E0 π 2 + ππΏ 2 π2 = π 2 + ππΏ 2 π = π 2 + ππΏ 2 E0 sin ππ‘ πΏ π ππ π ππΏ E0 π0
- 17. β | tan π = ππΏ π Steady state current (π) in the circuit π = E0 π sin(ππ‘ β π) E0 sin ππ‘ πΏ π ππ π ππΏ E0 π0 π = π 2 + ππΏ 2 π0 = E0 π = E0 π 2 + ππΏ 2 The current lags the emf by π.
- 18. In the given circuit, find (a) inductive reactance ππΏ . (b) impedance π. (c) Peak current π0. (d) i(t) π = 200 sin(100ππ‘) πΏ = 2 π π» π = 200 Ξ©
- 19. π = 200 sin(100ππ‘) πΏ = 2 π π» π = 200 Ξ© (a) inductive reactance ππΏ . ππΏ = ππΏ ππΏ = 100π Γ 2 π = 200 Ξ© π ππΏ π π π2 = π 2 + ππΏ 2 π = π 2 + ππΏ 2 β΄ π = 200 2 Ξ© π = (200)2+ 200 2 (b) impedance π.
- 20. π = 200 sin(100ππ‘) πΏ = 2 π π» π = 200 Ξ© (c) Peak current π0. π0 = π0 π π0 = 200 π π0 = 200 200 2 = 1 2 π΄ ππ π ππΏ Ξ΅0 π0 π ππΏ π π (d) i(t) π π‘ = π0 sin 100ππ‘ β π tan π = ππΏ π = 200 200 = 1 β π = π 4 β΄ π π‘ = 1 2 sin 100ππ‘ β π 4