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- 1. Welcome to Classes BYJU’S Alternating Current What you already know What you will learn S6: Power in AC circuits 1 . Pure AC circuits 2 . RC and L R AC circuits 3 . Im pe dance 1 . Total work done in a cy cle 2 . Av e rage po we r de liv e re d in pu re re sistiv e and pu re re activ e circu its 3 . Powe r in RC and RL circu its
- 2. Impedance The relation b/w peak current and peak voltages can be written as 𝑖0 = E0 𝑍 𝑍 = 𝑅 𝑍 = 𝜔𝐿 𝑍 is called impedance. Impedance is defined as the opposition any circuit presents when voltage is applied to it. SI unit is Ohm (Ω) 𝑍 = 1/𝜔𝐶
- 3. RC combination in AC circuits 𝑍 = 𝑅2 + 1/𝜔𝐶 2 tan 𝜙 = 1 𝜔𝐶𝑅 𝑖0 = E0 𝑍 = E0 𝑅2 + 1/𝜔𝐶 2 Steady state current (𝑖) in the circuit, 𝑖 = E0 𝑍 sin(𝜔𝑡 + 𝜙) E0 sin 𝜔𝑡 𝐶 𝑅 The current leads the emf by 𝜙. 𝑉𝑅 𝜙 𝑉𝐶 E0 𝑖0 𝜙 𝑋𝐶 𝑅 𝑍
- 4. LR combination in AC circuits tan 𝜙 = 𝜔𝐿 𝑅 Steady state current (𝑖) in the circuit 𝑖 = E0 𝑍 sin(𝜔𝑡 − 𝜙) E0 sin 𝜔𝑡 𝐿 𝑅 𝑉𝑅 𝜙 𝑉𝐿 E0 𝑖0 𝑍 = 𝑅2 + 𝜔𝐿 2 𝑖0 = E0 𝑍 = E0 𝑅2 + 𝜔𝐿 2 The current lags the emf by 𝜙. 𝜙 𝑋𝐿 𝑅 𝑍
- 5. A series R-C circuit is connected to an alternating voltage source. Consider two situations (a) When capacitor is air filled. (b) When capacitor is mica filled. Current through resistor is 𝐼 and voltage across capacitor is 𝑉 then- a b c 𝑉 𝑎 = 𝑉𝑏 𝑉 𝑎 < 𝑉𝑏 𝑉 𝑎 > 𝑉𝑏 𝐼𝑎 > 𝐼𝑏 d
- 6. 𝑉 𝑎 > 𝑉𝑏 𝑋𝐶 ∝ 1 𝐶 When capacitor is filled with mica, its capacitance increases. As 𝐶 increases, 𝑋𝐶 decreases As 𝑋𝐶 decreases, voltage across capacitor decreases (𝑋𝐶 ∝ 𝑉). 𝐶 𝐶′ 𝐶′ > 𝐶 𝑋𝐶 = 1 𝜔𝐶 a 𝑉 𝑎 = 𝑉𝑏 𝑉 𝑎 < 𝑉𝑏 𝑉 𝑎 > 𝑉𝑏 𝐼𝑎 > 𝐼𝑏 d c b
- 7. An A.C. voltage is applied to a resistance 𝑅 and an inductor 𝐿 in series. If 𝑅 and the inductive reactance are both equal to 3 Ω, the phase difference between the applied voltage and the current in the circuit is E0 sin 𝜔𝑡 𝑋𝐿 = 3 Ω 𝑅 = 3 Ω a b c d Zero 𝜋/6 𝜋/4 𝜋/2
- 8. tan 𝜙 = 𝑋𝐿 𝑅 tan 𝜙 = 𝑋𝐿 𝑅 = 1 𝜙 = 45° or 𝜋/4 E0 sin 𝜔𝑡 𝑋 = 3 Ω 𝑅 = 3 Ω 𝜙 𝑋𝐿 𝑅 𝑍 a b d Zero 𝜋/6 𝜋/4 𝜋/2 c
- 9. In an A.C. circuit, an alternating voltage, 𝜀 = 200 2 sin(100𝑡) volt is connected to capacitor of capacity 1 𝜇𝐹. The r.m.s. value of current in the circuit is 𝜀 = 200 2 sin(100𝑡) 1 𝜇𝐹 a b c d 10 𝑚𝐴 100 𝑚𝐴 200 𝑚𝐴 20 𝑚𝐴
- 10. Alternating voltage, ε = 200 2 sin(100𝑡) comparing with ε = 𝜀0 sin 𝜔𝑡 𝜔 = 100 𝑟𝑎𝑑/𝑠 𝜀0 = 200 2 𝑣𝑜𝑙𝑡 𝑋𝐶 = 1 𝜔𝐶 = 1 100 × 10−6 Ω = 104 Ω 𝐼0 = 𝜀0 𝑋𝐶 = 200 2 104 𝐴 = 2 2 × 10−2 𝐴 𝐼𝑟𝑚𝑠 = 𝐼0 2 = 2 2×10−2 2 = 2 × 10−2𝐴 = 20 𝑚𝐴 𝜀 = 200 2 sin(100𝑡) 1 𝜇𝐹 𝜀 = 𝜀0 sin 𝜔𝑡 a b c 10 𝑚𝐴 100 𝑚𝐴 200 𝑚𝐴 20 𝑚𝐴 c
- 11. E = E0 sin 𝜔𝑡 𝑖 = 𝑖0 sin(𝜔𝑡 + 𝜙) 𝑑𝑊 = E0𝑖0 sin 𝜔𝑡 sin(𝜔𝑡 + 𝜙) 𝑑𝑡 = E0𝑖0 sin2 𝜔𝑡 cos 𝜙 + sin 𝜔𝑡 cos 𝜔𝑡 sin 𝜙 𝑑𝑡 sin(𝜔𝑡 + 𝜙) Total work done in a cycle is, 𝑊 = E0𝑖0 cos 𝜙 0 𝑇 sin2 𝜔𝑡 𝑑𝑡 + E0𝑖0 sin 𝜙 0 𝑇 sin 𝜔𝑡 cos 𝜔𝑡 𝑑𝑡 = E0𝑖0 cos 𝜙 0 𝑇 (1 − cos 2𝜔𝑡) 𝑑𝑡 + E0𝑖0 sin 𝜙 0 𝑇 1 2 sin 2𝜔𝑡 𝑑𝑡 𝑊 = 1 2 E0𝑖0 𝑇 cos 𝜙 The work done by source in time interval 𝑑𝑡 is, 𝑑𝑊 = E 𝑖 𝑑𝑡 𝐿 𝐶 𝑅 = E0𝑖0 cos 𝜙 2 0 𝑇 𝑑𝑡 − 0 𝑇 cos 2𝜔𝑡 𝑑𝑡 + E0𝑖0 sin 𝜙 0 𝑇 1 2 sin 2𝜔𝑡 𝑑𝑡
- 12. Total work done in a cycle is, 𝑊 = 1 2 E0𝑖0 𝑇 cos 𝜙 Average Power delivered, 𝑃𝑎𝑣𝑔 = 𝑊 𝑇 = E0 2 𝑖0 2 cos 𝜙 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 𝑃𝑎𝑣𝑔 = 1 2 E0 𝑖0 cos 𝜙 E = E0 sin 𝜔𝑡 𝑖 = 𝑖0 sin(𝜔𝑡 + 𝜙) 𝐿 𝐶 𝑅
- 13. E = E0 sin 𝜔𝑡 𝑖 = 𝑖0 sin(𝜔𝑡 + 𝜙) 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 Power Factor For purely resistive circuit, 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 0° 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 Power drawn is maximum in a purely resistive circuit 𝑅 𝜀 = 𝜀0 sin 𝜔𝑡 𝑡 𝑖, 𝜀 0 𝜋 2𝜋 𝜋 2 3𝜋 2 𝜀0 𝑖0
- 14. 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 Power Factor For purely reactive circuit - 𝑡 𝑖, 𝜀 0 𝜋 2𝜋 𝜋 2 3𝜋 2 𝜀 = 𝜀0 sin 𝜔𝑡 𝑖 = 𝑖0 sin 𝜔𝑡 − 𝜋 2 𝜀0 𝑖0 𝑖0 𝑡 𝑖, 𝜀 0 𝜋 2𝜋 𝜋 2 3𝜋 2 𝜀 = 𝜀0 sin 𝜔𝑡 𝜀0 𝑖0 𝑖0 𝑖 = 𝑖0 sin 𝜔𝑡 + 𝜋 2 𝜙 = 𝜋/2 𝜙 = −𝜋/2 𝜙 = 𝜋/2 or 𝜙 = −𝜋/2 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos ± 𝜋 2 𝑃𝑎𝑣𝑔 = 0 No power is absorbed for a full cycle in purely inductive or purely capacitive circuits
- 15. | 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 E0 sin 𝜔𝑡 𝐶 𝑅 cos 𝜙 = 𝑅 𝑍 cos 𝜙 = 𝑅 𝑅2 + 1/𝜔𝐶 2 Average Power in RC circuits is, 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠𝑖𝑟𝑚𝑠 𝑅 𝑍 𝜙 𝑋𝐶 𝑅 𝑍
- 16. 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 cos 𝜙 = 𝑅 𝑍 cos 𝜙 = 𝑅 𝑅2 + 𝜔𝐿 2 Average Power in LR circuits is, 𝑃𝑎𝑣𝑔 = E𝑟𝑚𝑠𝑖𝑟𝑚𝑠 𝑅 𝑍 E0 sin 𝜔𝑡 𝐿 𝑅 E0 sin 𝜔𝑡 𝜙 𝑋𝐿 𝑅 𝑍 |