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![Series L-C-R circuit 𝑅
𝑉 = 𝑉𝑚 sin(𝜔𝑡)
𝐶
𝐿
𝐼 = 𝐼𝑚 sin(𝜔𝑡 + 𝜙)
𝐼
𝑉𝑅
𝜔𝑡 + 𝜙
𝑉𝐿
𝑉𝐶
𝑉𝑅
𝑉𝐶𝑚 − 𝑉𝐿𝑚
𝜙
𝜔𝑡
𝑉
𝑉𝑚
2 = 𝑉𝑅𝑚
2
+ 𝑉𝐶𝑚 − 𝑉𝐿𝑚
2
𝑉𝑚
2 = 𝐼𝑚 𝑅 2 + 𝐼𝑚 𝑋𝐶 − 𝐼𝑚 𝑋𝐿
2
𝑉𝑚
2 = 𝐼𝑚
2 [𝑅2 + 𝑋𝐶 − 𝑋𝐿
2]
𝐼𝑚 =
𝑉𝑚
𝑅2 + 𝑋𝐶 − 𝑋𝐿
2
𝑉𝐶 + 𝑉𝐿
𝑉𝑅𝑚](https://image.slidesharecdn.com/resonance-240101135029-53b890be/75/Resonance-pdf-4-2048.jpg)
Resonance.pdf
- 1. Welcome to Classes BYJU’S Alternating Current Whatyou already know What you will learn S8: Resonance 1 . Pure AC circuits 2 . RC and L R AC circuits 3 . Impedance 4. Powe r in AC circuits 5 . Se rie s L CR circuits 1 . Condition for re sonance 2 . Graphical representation 3 . Num e ricals on re sonance
- 2. Apparent power Total powerflowing in a circuit is known as apparent power. 𝑆 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 Reactive power is product of voltage and current off phase by 𝜋 2 with the voltage 𝑄 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 sin 𝜙 Reactive power Current out of phase Current in phase E 𝐼 𝐼 sin 𝜙 𝜙 𝐼 cos 𝜙 𝑃 = E𝑟𝑚𝑠 𝑖𝑟𝑚𝑠 cos 𝜙 Active power is product of voltage and current in phase with the voltage. Active power
- 3. Power factor The ratioof Active power to Apparent power is the power factor. 𝑃 𝑆 = cos 𝜙 Relationship between apparent, reactive and active power E 𝑆 𝑄 = E𝐼 sin 𝜙 𝜙 𝑃 = E𝐼 cos 𝜙 E 𝐼 𝐼 sin 𝜙 𝜙 𝐼 cos 𝜙 Where, 𝑆 = Apparent power 𝑄 = Reactive power 𝑃 = Active power 𝑆 = 𝑃2 + 𝑄2
- 4. Series L-C-R circuit𝑅 𝑉 = 𝑉𝑚 sin(𝜔𝑡) 𝐶 𝐿 𝐼 = 𝐼𝑚 sin(𝜔𝑡 + 𝜙) 𝐼 𝑉𝑅 𝜔𝑡 + 𝜙 𝑉𝐿 𝑉𝐶 𝑉𝑅 𝑉𝐶𝑚 − 𝑉𝐿𝑚 𝜙 𝜔𝑡 𝑉 𝑉𝑚 2 = 𝑉𝑅𝑚 2 + 𝑉𝐶𝑚 − 𝑉𝐿𝑚 2 𝑉𝑚 2 = 𝐼𝑚 𝑅 2 + 𝐼𝑚 𝑋𝐶 − 𝐼𝑚 𝑋𝐿 2 𝑉𝑚 2 = 𝐼𝑚 2 [𝑅2 + 𝑋𝐶 − 𝑋𝐿 2] 𝐼𝑚 = 𝑉𝑚 𝑅2 + 𝑋𝐶 − 𝑋𝐿 2 𝑉𝐶 + 𝑉𝐿 𝑉𝑅𝑚
- 5. Impedance 𝑅 𝑋𝐶 𝑋𝐿 𝑉𝑅𝑚 𝐼 𝑉𝐶𝑚 𝑉𝐿𝑚 𝑉𝐿𝑚 −𝑉𝐶𝑚 𝑉𝑅𝑚 𝐼 𝑉𝑚 𝜙 𝜙 𝑅 𝑋𝐿 − 𝑋𝐶 𝑍 tan 𝜙 = 𝑋𝐿 − 𝑋𝐶 𝑅 tan 𝜙 = 𝑉𝐿𝑚 − 𝑉𝐶𝑚 𝑉𝑅𝑚 Case 1 : 𝑉𝐿𝑚 > 𝑉𝐶𝑚 or 𝑋𝐿 > 𝑋𝐶
- 6. Impedance 𝑅 𝑋𝐶 𝑋𝐿 𝑉𝑅𝑚 𝐼 𝑉𝐶𝑚 𝑉𝐿𝑚 Case 2: 𝑉𝐿𝑚 < 𝑉𝐶𝑚 or 𝑋𝐿 < 𝑋𝐶 𝑉𝐶𝑚 − 𝑉𝐿𝑚 𝑉𝑅𝑚 𝜙 𝜙 𝑅 𝑋𝐶 − 𝑋𝐿 𝑍 tan 𝜙 = 𝑋𝐶 − 𝑋𝐿 𝑅 tan 𝜙 = 𝑉𝐶𝑚 − 𝑉𝐿𝑚 𝑉𝑅𝑚 𝑉𝑚
- 7. In series LCRcircuit, resistance 𝑅 = 10 Ω and impedance 𝑍 = 20 Ω. The phase difference between the current and the voltage is: a b c 30° 45° 60° 90° d
- 8. 𝜙 𝑅 |𝑋𝐿 − 𝑋𝐶| 𝑅 = 10 Ω 𝐶 𝐿 𝐴 cos 𝜙 = 𝑅 𝑍 = 10 20 = 1 2 𝑍 = 20 Ω 𝜙 = cos−1 0.5 = 60° c a b 30° 45° 60° 90° d
- 9. In a circuit,𝐿, 𝐶 and 𝑅 are connected in series with an alternating voltage source of frequency 𝑓. The current leads the voltage by 45° . The value of 𝐶 is a b c 1 2ߨ݂(2ߨ݂ܮ+ܴ) 1 2𝜋𝑓(2𝜋𝑓𝐿 − 𝑅) 1 𝜋𝑓(2𝜋𝑓𝐿 − 𝑅) 1 𝜋𝑓 2𝜋𝑓𝐿 + 𝑅 d
- 10. tan 𝜙 = 𝑋𝐶− 𝑋𝐿 𝑅 (∵ Current leads the source voltage) tan 45∘ = 1 2𝜋𝑓𝐶 − 2𝜋𝑓𝐿 𝑅 𝑅 = 1 2𝜋𝑓𝐶 − 2𝜋𝑓𝐿 ⇒ 2𝜋𝑓𝐶 = 1 2𝜋𝑓𝐿 + 𝑅 ∴ 𝐶 = 1 2𝜋𝑓 2𝜋𝑓𝐿 + 𝑅 𝑅 𝑓 𝐻𝑧 𝐶 𝐿 b c d a 1 2ߨ݂(2ߨ݂ܮ+ܴ) 1 2𝜋𝑓(2𝜋𝑓𝐿 − 𝑅) 1 𝜋𝑓(2𝜋𝑓𝐿 − 𝑅) 1 𝜋𝑓 2𝜋𝑓𝐿 + 𝑅
- 11. CASE 2 If 𝑋𝐶< 𝑋𝐿 , 𝜙 ≠ 0 Circuit is predominantly inductive Current lags the source voltage CASE 1 If 𝑋𝐶 > 𝑋𝐿 , 𝜙 ≠ 0 Circuit is predominantly capacitive Current leads the source voltage CASE 3 if 𝑋𝐶 = 𝑋𝐿 , 𝜙 = 0 𝒕𝒂𝒏 𝝓 = 𝑿𝑳 − 𝑿𝑪 𝑹 𝜙 𝑅 |𝑋𝐿 − 𝑋𝐶 |
- 12. 𝑅 𝐶 𝐿 𝑖𝑚 = 𝑉𝑚 𝑅2 +𝑋𝐶 − 𝑋𝐿 2 𝑋𝐶 = 1 𝜔𝐶 𝑋𝐿 = 𝜔𝐿 If 𝜔 is varied, then at a particular frequency (𝜔0), 𝑋𝐶 = 𝑋𝐿 𝑉 = 𝑉𝑚 sin(𝜔𝑡)
- 13. 𝑖𝑚 = 𝑉𝑚 𝑅2 +𝑋𝐶 − 𝑋𝐿 2 𝑋𝐶 = 1 𝜔𝐶 𝑋𝐿 = 𝜔𝐿 𝜔0 is resonant angular frequency Impedance is minimum (𝑍 = 𝑅2 + 02 = 𝑅) and purely resistive circuit Current is maximum (𝑖𝑚 = 𝑉𝑚 /𝑅) If 𝜔 is varied, then at a particular frequency (𝜔0), 𝑋𝐶 = 𝑋𝐿
- 14. 𝑋𝐶 = 1 𝜔𝐶 𝑋𝐿 =𝜔𝐿 For resonance condition, 𝑋𝐶 = 𝑋𝐿 1 𝜔0𝐶 = 𝜔0𝐿 𝜔0 = 1 𝐿𝐶 𝑓0 = 1 2𝜋 𝐿𝐶 Resonant frequency 𝑅 𝐶 𝐿 𝑉 = 𝑉𝑚 sin(𝜔𝑡)
- 15. | 𝑓 𝑓0 𝑋𝐶 = 1 2𝜋𝑓𝐶 𝑋𝐿 =2𝜋𝑓𝐿 𝑋𝐿 𝑅 𝑋𝐶 Independent of frequency Resonance condition, 𝑋𝐶 = 𝑋𝐿 𝑋𝐿 , 𝑋𝐶 , 𝑅
- 16. | 𝑓 𝑓0 𝑋𝐿 𝑅 𝑋𝐶 𝑍 (Ω) 𝑓 𝑓0 𝑍𝑚𝑖𝑛 = 𝑅 𝑍= 𝑅2 + 1 2𝜋𝑓𝐶 − 2𝜋𝑓𝐿 2 Resonance condition, 𝑋𝐶 = 𝑋𝐿 Capacitive, 𝑋𝐶 > 𝑋𝐿 Inductive, 𝑋𝐶 < 𝑋𝐿
- 17. 𝑖 𝑚 (𝐴 ) 𝜔 (𝑟𝑎𝑑/𝑠) 𝜔0 𝑍 =𝑅2 + 1 𝜔𝐶 − 𝜔𝐿 2 𝑅 𝐶 𝐿 𝑉 = 𝑉𝑚 sin(𝜔𝑡) 𝑖𝑚 = 𝑉𝑚 /𝑍 𝑖𝑚𝑎𝑥 1 2 3 𝑅3 > 𝑅2 > 𝑅1
- 18. What is thevalue of inductance 𝐿 for which the current is maximum in a series LCR circuit with 𝐶 = 10 𝜇𝐹 and 𝜔 = 1000 𝑠−1? a b c 1 𝑚𝐻 10 𝑚𝐻 100 𝑚𝐻 Cannot be calculated unless 𝑅 is known d
- 19. 𝑋𝐿 = 𝑋𝐶⇒ 𝜔𝐿 = 1 𝜔𝐶 𝐿 = 1 𝜔2𝐶 = 1 1000 2 × 10 × 10−6 𝐿 = 1 10 = 0.1 𝐻 = 100 𝑚𝐻 For maximum current in series 𝐿𝐶𝑅 circuit, a b d c 1 𝑚𝐻 10 𝑚𝐻 100 𝑚𝐻 Cannot be calculated unless 𝑅 is known
- 20. a b c 𝜋𝑍𝑒𝑟𝑜 𝜋/4 𝜋/2 d A LCR circuit is connected to a source of alternating current. At resonance, find the phase difference between the applied voltage and the current in the circuit.
- 21. 𝐼 𝑉𝑅 𝜔𝑡 + 𝜙 𝑉𝐿 𝑉𝐶 Inresonance condition, 𝑉𝐿 = 𝑉𝐶 ∴ Phase difference between applied voltage and current = 0 a c d b 𝜋 𝑍𝑒𝑟𝑜 𝜋/4 𝜋/2
- 22. A transistor-oscillator usinga resonant circuit with an inductor 𝐿 (of negligible resistance) and a capacitor 𝐶 in series produce oscillation of frequency 𝑓. If 𝐿 is doubled and 𝐶 is changed to 4𝐶, the frequency will be a b c 𝑓/2 2 𝑓/2 𝑓/4 8𝑓 d
- 23. 𝑓 = 1 2𝜋 𝐿𝐶 𝑓= 𝑓1 = 1 2𝜋 𝐿1𝐶1 = 1 2𝜋 𝐿𝐶 𝑓2 = 1 2𝜋 𝐿2 𝐶2 = 1 2𝜋 2𝐿 × 4𝐶 𝑓2 𝑓1 = 𝐿𝐶 8𝐿𝐶 = 1 2 2 𝑓2 = 𝑓 2 2 a b c 𝑓/2 2 𝑓/2 𝑓/4 8𝑓 d