Title 07. ALTERNATING CURRENT(H).pdf ✅

NEET REVISION 07. ALTERNATING CURRENT(H) NEET REVISION Date: March 18, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on The voltage and are equal and opposite so voltmeter reading will be zero. Also 2. () : Explana on For so the circuit is in resonance and phase difference between current and voltage is zero. For current leads the voltage For voltage leads the current. So op on (1) is correct. 3. () : Explana on so bulb glows brighter then 4. () : Explana on As if increased, then decreases. As increases, will be increases. will be same as it not a func on of . Total impedance will be minimum at resonance. 5. () : Explana on The equa on of the graph from 0 to is The average value is 6. () : Explana on VL VC R = 30Ω, XL = XC = 25Ω ⇒ i = = = = 8 A V √R2+(XL−XC) 2 V R 240 30 ω = 5000 rad/ sec XL = ωL = 5000 × 4 × 10 −3 = 20Ω XC = = = 20Ω 1 ωC 1 5000×10×10−2 XL = XC ω = 2500 rad/ sec XL = ωL = 2500 × 4 × 10 −3 = 10Ω XC = = = 40Ω 1 ωC 1 2500×10×10−6 XC > XL ω = 10000 rad/ sec XL = 10 4 × 4 × 10 −3 = 40Ω XC = = 10Ω 1 10 4×10×10−6 XL > XC XC = = = Ω 1 ωC 10 6 100π×100 20 π XL = ωL = 100π × 10 −2 = πΩ XC > XL ⇒ XRC > XLC i2 > i1 B2 B1 XC = 1 ωC ⇒ XC ∝ 1 ω ⇒ ω XC ⇒ (A − S) XL = ωL ⇒ XL ∝ ω ⇒ (B − P) ⇒ ω XL R ω ⇒ (C − Q) ⇒ (D − R) π V = Vm sin θ Vavg = ∫ π 0 V dθ π Vavg = ∫ π 0 sin θdθ = V 2 eff = ∫ π 0 sin 2 θdθ = Veff = Vm/√2 Vm π 2Vm π V 2m π V 2m 2 XL = ωL = 500 × 0.02 = 10Ω i0 = = 10A i = 10 sin[ωt − π/2] = −10 cos 500t V0 XL

NEET REVISION 17. () : Explana on Given But 18. () : Explana on , Current in primary coil 19. () : Explana on 20. () : Explana on In circuit (1), on closing the switch, the current in the inductor is zero due to selfinduc on, . In circuit (2), on closing the switch, the current in the inductor is zero dut to self-induc on. Therefore, In circuit (3), on closing the switch, the current in the inductor is again zero due to the same reason. Therefore, Thus, it is obvious that, . 21. () : Explana on Here, The current in the coil is The energy stored in the coil is 22. () : Explana on Phase angle between and is . 23. () : Explana on Since the circuit is at resonance, the current in the circuit is in the phase with applied voltage. Voltage across inductor leads the current by and across a capacitor lags by . So, the voltage across resistance is lagging by than the voltage across capacitor. 24. () : Explana on I = 2 + 5 sin ωt Irms = ⎡ ⎣ ⎤ ⎦ = [ ∫ T 0 (2 + 5 sin ωt) 2dt] 1/2 = [ ∫ T 0 (4 + 25 sin 2 ωt + 20 sin ωt) dt] 1/2 ∫ T 0 I 2dt ∫ T 0 dt 1 T 1 T ∫ T 0 sin ωtdt = 0 Irms = 4T + ∫ T 0 sin 2 ωtdt = 4 + ( ) = 16.5 1 T 25 T 25 T T 2 EP = 240 V , ES = 24 V ESIS = 24 W IP = = = 0.1 A ESIS EP 24 240 V = √V 2 R + (VL − VC) 2 = √40 2 + (40 − 10) 2 = √40 2 + 30 2 = 50 volt z = = = 5 Ω Vrms Irms 50 ( ) 10√2 √2 i. e. i1 = 0 i2 = i = E R1 i3 = i = E R1 + R2 i2 > i3 > i1(= 0) R = 10 Ω,L = 5H, V = 100 V I = = = 10 A V R 100 V 10 Ω U = LI 2 = (5H)(10 A) 2 = 250 J 1 2 1 2 V I 90 ∘ < P >= VrmsIrms cos = 0 π 2 π/2 π/2 90 ∘ Leq = L + L = 2L Ceq = = C1C2 C1+C2 C 2 f = = = 1 2π√LeqCeq 1 2π√2L× L 2 1 2π√LC
NEET REVISION 25. () : Explana on 26. () : Explana on Efficiency 27. () : Explana on We know of of Hence 28. () : Explana on The current through the resistor is 29. () : Explana on 30. () : Explana on As the current lags the emf by , it is an circuit. 31. () : Explana on For a purely induc ve circuit 32. () : Explana on For delivering maximum power from the generator to the passive load, total reactance must vanish, 33. () : Explana on We have, This condi on is sa sfied only when current in pri‐ mary is and in secondary 34. () : Explana on In the given circuit diagram, both inductors of inductance are parallel to each other, hence Similarly, Reasonance frequency 35. () : Explana on Conceptual Ques on ω = 2ωo = 2 × = 200 rads 1 −1 √LC XL = ωL = 200 × 5 = 1000Ω XC = = = 250Ω 1 ωC 1 200×20×10−6 I = = 5×10 −3 √(XL−XC) 2 5×10 −3 √(1000−250) 2 = = × 10 −4 A 5×10−3 750 1 15 VC = IXCandQ = CVC = 33.3 nC η = 0.9 = PS PP VSIS = 0.9 × VP IP IS = = 45 A 0.9×2300×5 230 v0 = 1 2π√LC L′ = L + 25% L = L + = ; L 4 5L 4 C′ = C − 20% C = C − = C 5 4C 5 v ′ 0 = = 1 2π√L′C′ 1 2π√ × 5L 4 4C 5 = = v0 1 2π√LC R = = 240Ω (120) 2 60 i = = = 0.5A P V 60W 120V Z = = = 480Ω Vrms irms 240 0.5 XL = √480 2 − 240 2 = 240√3Ω L = = H 240√3 60×2π 2√3 π ω = 2000π; XC = = 1 2000π × 10 −8 10 5 2π Z = √X2 C + R2 ≃ 10 5 2π ∴ VC ≃ Vin = 10mV i e π/4 L − R tan φ = ortan = XL R π 4 ωL R ∴ R = ωL As ω = 100rad/s If L = 10H then R = 1 KΩ P = 0 i.e., XL + Xg = 0 or XL = −Xg = , = ⇒ = ⇒ Is = 50Ip Ns Np Ip Is 10 500 Ip Is Ip Is 1 50 3.2 × 10 −3 A 0.16 A. L Leg = = = L1L2 L1+L2 L⋅L L+L L 2 Ceg = C1 + C2 = C + C = 2C ∴ ωr = = √ 1 √Leg . Ceg 1 . 2C L 2 = √ = 1 LC 1 √LC