Tiêu đề 07. ALTERNATING CURRENT.pdf ✅

07. ALTERNATING CURRENT NEET PREPARATION (MEDIUM PHYSICS PAPER) Date: March 12, 2025 Dura on: 1:00:00 Total Marks: 180 INSTRUCTIONS INSTRUCTIONS PHYSICS 1. () : Explana on 2. () : Explana on 3. () : Explana on In parallel resonance circuit, impedance At resonance, then The impedance is maximum Current is zero (or) minimum At resonance 4. () : Explana on So, correct op on is (2). 5. () : Explana on 6. () : Explana on Power factor, When is doubled, we find 7. () : Explana on Let is the angle made by w.r.to The Wa less current is 8. () : Explana on This is because the reason provides a valid ex‐ plana on for why knowing the power factor alone does not determine whether the e.m.f. leads or lags the current in an circuit. For that, we shall know whether (or) . So correct op on is (1). = ⇒ = Np NS IS IP 1 25 2 IP IP = 50 A P = V0 i0 cos φ ⇒ 1000 = × 200 × i0 cos 60 1 ∘ 2 1 2 ⇒ i0 = 20A ⇒ irms = = = 10√2A. i0 √2 20 √2 LC (Z) = XC,XL XC−XL XC = XL Z → ∞ ⇒ ⇒ XL = XC ωL = 1 ωC ω = 1 √LC Z = √R2 + (XL − XC) 2 ⇒ |Z| = √5 + 4 = 3Ω XC = 1 2πvC ⇒ C = = = 50μF 1 2πvXC 1 2×π× ×25 400 π cos φ = = ⇒ ω 2L 1 2 = R2 √2 R √R2+ω2L2 ω cos φ ′ = = = R √R2+4ω2L2 R √R2+4R2 1 √5 φ V i irms sin φ iWL = irms sin φ ⇒ √3 = 2 sin φ ⇒ sin φ = √ 2 ⇒ θ = 60 ∘ ⇒ cos φ = cos 60 ∘ = 1 2 LCR XL > XC XC > XL

21. () : Explana on The voltmeter connected to mains is cali‐ brated to read root mean square value of voltage, i.e, . 22. () : Explana on Here, The maximum current is Maximum poten al difference is 23. () : Explana on Alterna ng current , where If the current, star ng from zero, reaches rms value at me Similarly if the current reaches peak value at me Hence me to reach peak value from rms value: 24. () : Explana on Ammeter reads the root mean square value of current is related to the peak value of current by the rela on 25. () : Explana on Phase difference between voltage and current Power factor Average power dissipated 26. () : Explana on Reasonant frequency in an series circuit is To reduce can be increased, by adding an‐ other capacitor in parallel to the first capacitor 27. () : Explana on A step-down transformer converts electrical en‐ ergy from a high voltage to one at a low voltage. (Subscript for primary and for secondary) constant In order to produce less heat in secondary, we shall use a wire of less resistance and so "thick wire" ( high in of formule radia on, will be maximum when plane of armature is parallel to lines of force of mag‐ ne c field. (and then in maximum then).Both are facts, but asser on is true for a different explana on. So correct op on is (2). 28. () : Explana on Given, and Now, maximum poten al difference 29. () : Explana on voltmeter reads rms value of voltage. RMS Voltage applied Reading of voltmeter is , then AC AC √< v 2 > irms = 10A, R = 12Ω Im = √2Irms = √2(10) = 10√2A Vm = ImR = 10√2 × 12 = 169.68V . I = I0 sin(2πvt) v = 50 Hz t1, I0/√2 = I0 sin (2πvt1) ⇒ sin (2πvt1) = 1/√2 = sin ( ) π 4 ∴ t1 = 1/(8v) t2 , I0 = I0 sin (2πvt2) ⇒ sin (2πvt2) = 1 = sin(π/2) ∴ t2 = 1/(4v) t2 − t1 = 1/(8v) = 1/(8 × 50)s = 1/400 s = 2.5 × 10 −3s (Irms) (I0) Irms = ⇒ I0 = √2 × Irms = 20√2A I0 √2 φ = − (− ) = π 12 π 6 π 4 ⇒ = cos φ = 1 √2 = cos φ VmIm 2 = × = 5√2 watt 10×2 2 1 √2 L − C − R υr = 1 2π√LC υr; C p s ⇒ Vs < Vp ⇒ Is > Ip(∵ P = V I = ) i. e. , r A = πr 2 R = ρ ) l A ( ) dφ dt |Vind | Irms = 10 A R = 12 Ω Erms = Irms × R = 10 × 12 = 120 V = Erms × √2 = 120 × √2 = 169.68 V AC = = 50V Vo √2 V2 x √x 2 + 40 2 = 50V ⇒ x = 30V