Tiêu đề 7.ALTERNATING CURRENT - Explanations.pdf ✅

2 (b) The impedance of R – C circuit for frequency f1 is Z1 = √R2 + 1 4π2f 2C 2 The impedance of R – C circuit for frequency 2f is Z2 = √R2 + 1 4π2(2f 2)C 2 or Z2 = √R2 + 1 16π2f 2C 2 Then, Z1 2 Z2 2 = R 2+ 1 4π2f2C2 R2+ 1 16π2f2C2 Or Z1 2 Z2 2 = 1+ 1 4π2f2C2R2 1+ 1 16π2f2R2C2 Values are greater than 1 then Z1 Z2 = lies between 1 and 2. 4 (b) The coil has inductance L besides the resistance R. Hence for ac it’s effective resistance √R2 + XL 2 will be larger than it’s resistance R for dc 5 (a) At resonance LCR series circuit behaves as pure resistive circuit. For resistive circuit φ = 0° 6 (c) Heat produced by ac = 3 × Heat produced by dc ∴ irms 2 Rt = 3 × i 2 Rt ⇒ irms 2 = 3 × 2 2 ⇒ irms = 2√3 = 3.46 A 7 (c) Power =Rate of work done in one complete cycle. or Pav = W T or Pav = (E0I0 cos φ)T/2 T or Pav = E0I0 cosφ 2 Where cosφ is called the power factor of an AC circuit. 8 (c) The circuit element connected to the AC source will be pure resistor. In pure resistive AC circuit, voltage and current are in the same phase 9 (b) As e = MdI/dt, ∴ M = edt dI = − 15000×0.001 3 = 5H 10 (a) Vrms = √ 1 T ∫ 102 dt T 0 = 10 V 11 (b) In non resonant circuits Impedance Z = 1 √ 1 R2+(ωC− 1 ωL) 2 , with rise in frequency Z decreases, i. e., current increases so circuit behaves as capacitive circuit 12 (b) Z = √R2 + XC 2 ∶ Irms = Vrms Z :P = Irms 2 R Where XC = 1 ωC As ω is increased, XCwill decrease or Z will decrease. Hence Irms or P will increase. Therefore, bulb glows brighter. Hence the correct option is (b). 13 (d) Phase angle φ = 90°, so power P = Vi cos φ = 0 15 (c) i = i0 (1 − e − Rt L ) ⇒ di dt = d dt i0 − d dt (i0e − Rt L ) = 0 + i0R L e − Rt L Initially, t = 0 ⇒ di dt = i0 × R L = E L = 5 2 = 2.5 As −1 16 (c) Amplitude of ac = i0 = V0 R = ωNBA R = (2πv)NB(πr 2) R ⇒ i0 = 2π × 200 60 × 1 × 10−2 × π × (0.3) 2 π 2 = 6 mA 17 (c) VL = 46 volts, VC = 40 volts, VR = 8 volts E.M.F. of source V = √8 2 + (46 − 40) 2 = 10 volts 18 (c) f = 1 2π√LC ∴ f = 1 2 × 3.14√0.5 × 10−3 × 20 × 10−6 = 1 2 × 3.14 × 10−4 ≈ 1600 Hz 19 (d) Current lags the voltage if ωL > 1 ωC f > 1 2π√LC ⇒ f > fr 20 (c) Vrms = V0 √2 = 770√2 2 ≈ 500 V 21 (c) Phase difference ∆φ = φ2 − φ1 = π 6 − ( −π 6 ) = π 3 22 (b) For purely capacitive circuit e = e0 sin ωt i = i0 sin (ωt + π 2 ) , i. e., current is ahead of emf by π 2 23 (c) In L − R circuit, current at any time t is given by
i = E R (1 − e − R L t ) = E R − E R e − R L t di dt = E R e − R L t ( R L ) = E L e − R L t Induced emf = L di dt = Ee − R L t From Eq. (i), iR = E − Ee − R L t Using Eq. (ii), iR = E − e or e = E − iR Therefore, graph between e and i is a straight line with negative slope and positive intercept. The choice (c) is correct. 24 (d) When resonance occurs emf E and current i are in phase. In this case, the impedance is minimum and current is maximum. At resonance inductive reactance is equal to capacitive reactance XL = XC 25 (a) The current is I = E0 √R2+ω2L 2 = 4 √4 2+(1000×3×10−3)2 = 0.8 A 26 (d) As the magnetic field directed into the paper is increasing at a constant rate, therefore, induced current should produce a magnetic field directed out of the paper. Thus current in both the loops must be anti-clock-wise. As area of loop on right side is more, therefore, induced emf o right side of loop will be more compared to the emf induced on the left-side of the loop [∴ e = − dφ dt = −A dB dt ] 28 (d) l1 = F R1 = 12 2 = 6 A E = L dl2 dt + R2 × l2 I2 = I0(1 − e −t/tc) ⇒ I0 = E R2 = 12 2 = 6A tc = L R = 400 × 10−3 2 = 0.2 I2 = 6(1 − e −t/0.2 ) Potential drop across L = E − R2L2 = 12 − 2 × 6(1 − e −bt) = 12e −5t 29 (b) Capacitive reactance is given by XC = 1 ω C Where C is capacitance and ω the angular frequency (ω = 2πf). ∴ XC = 1 2πfC ⇒ XC ∝ 1 f Hence, when frequency f increases capacitive reactance decreases. 30 (d) For a series L – C – R circuit at resonance Phase difference, φ = 0° Power factor= cos φ = 1 32 (b) As e = M di dt ∴ M = e di/dt = 25×10−3 15.0 = 1.67 × 10−3H As φ = Mi ∴ φ = 1.67 × 10−3 × 3.6 = 6 × 10−3 Wb = 6 m Wb 33 (b) Average power dissipated in an AC circuit Pav = VrmsIrms cosφ ...(i) Where the term cos φ is known as power factor. Given, Vrms = 100 V, R = 100 Ω, φ = 30° ∴ Irms = Vrms R = 100 100 = 1 A Putting the values in Eq. (i), we get Pav = 100 × 1 × cos 30° = 100 √3 2 = 50√3 = 86.6 W 34 (b) e = −LdI/dt = −5 × (−2) = +10 V 35 (d) In purely inductive circuit voltage leads the current by 90° 36 (b) Q – factor = 1 R √ L C = 1 6 √ 1 17.36×10−6 = 40 37 (a) Resistance , R = 100 10 = 10 Ω Inductive reactance , XL = 2πfL 100 8 = 2π × 50 × L ⇒ L = 1 8π H XL ′ = 2πf ′L = 2π × 40 × 1 8π = 10 Ω Impedance of the circuit is Z = √R2 + XL ′ 2 = √(10) 2 + (10) 2 = 10√2 Ω Current in the circuit is i = V Z = 150 10√2 = 15 √2 A 38 (c)
ns nP = Es EP = 2400 120 = 20 ns = 20 nP = 20 × 75 = 1500. 39 (c) In a series L – C – R circuit, potential difference leads the current by an angle φ(let). φ = tan−1 ( XL−XC R ) or φ = tan−1 ( ωL− 1 ωC R ) At resonance, XL = XC, ie, ωL = 1 ωC Hence, φ = tan−1(0) = 0 Therefore, phase difference between current and voltage at resonance is zero. 40 (c) Here: Current in the circuit (i) = 15 mA = 15 × 10−3A Resistance R = 4000 Volt Applied voltage in the circuit = 240 V At any instant, the emf of the battery is equal to the sum of potential drop on the resistor and the emf developed in the induction coil Hence, E = iR + L di dt 240 = 15 × 10−3 × 4000 + L di dt Hence, L di dt = 240 − 60 = 180 V 41 (b) When C is removed circuit becomes RL circuit hence tan π 3 = XL R ... (i) When L is removed circuit becomes RC circuit hence tan π 3 = XC R ... (ii) From equation (i) and (ii) we obtain XL = XC. This is the condition of resonance and in resonance Z = R = 100Ω 42 (a) Induced emf produced in coil e = −dφ dt = −d dt (BA) ∴ |e| = A dB dt = 0.01 × 1 1 × 10−3 |e| = 10 V Current produced in coil, i = |e| R = 10 2 = 5 A Heat evolved = i 2Rt = (5) 2 × (2) × 1 × 10−3 = 0.05 J 43 (b) V0 = i0Z ⇒ 200 = 100 Z ⇒ Z = 2Ω Also Z 2 = R 2 + XL 2 ⇒ (2) 2 = (1) 2 + XL 2 ⇒ XL = √3Ω 44 (d) In an AC circuit, the coil of high inductance and negligible resistance used to control current, is called the choke coil. The power factor of such a coil is given by Cosφ = R √R2+ω2L 2 ≈ R ωL (as R << ωL) As R << ωL, cos φ is very small. Thus, the power absorbed by the coil is very small. The only loss of energy is due to hysteresis in the iron core, which is much less than the loss of energy in the resistance that can also reduce the current if placed instead of the choke coil. 45 (d) As a given pole (N or S) of suspended magnet goes into the coil and comes out of its, current is induced in the coil in two opposite directions. Therefore, galvanometer deflection goes to left and right both. As amplitude of oscillation of magnet goes on decreasing, so does the amplitude of deflection. 46 (a) Impedance, Z = √R2 + (XL − XC ) 2 ∴ 10 = √(102 + (XL − XC ) 2 ⇒ 100 = 100 + (XL − XC ) 2 ⇒ XL − XC = 0 ...(i) Let φ is the phase difference between current and voltage tanφ = XL−XC R ∴ tanφ = 0 R ⇒ φ = 0 [From Eq.(i)] 47 (a) XC = 1 2πvC ⇒ C = 1 2πvXC = 1 2 × π × 400 π × 25 = 50 μF 48 (b) ir.m.s. = i0 √2 = 4 √2 = 2√2 ampere 51 (b) P = Vi cosφ = V ( V Z ) ( R Z ) = V 2R Z 2 = V 2R (R2 + ω2L 2) 52 (c) 2πv = 377 ⇒ v = 60.03 Hz 53 (b) At t = 0, phase of the voltage is zero, while phase of the current is − π 2 , i. e., voltage leads by π 2 54 (c) The full cycle of alternating current consists of two half cycles. For one – half, current is positive
and for second half, current is negative. Therefore, for an AC cycle, the net value of current average out to zero. While the DC ammeter, read the average value. Hence, the alternating current cannot DC measured by DC ammeter. 55 (b) Ist case From formula R = V 2 P = 110×110 330 = 110 3 Ω Since, current lags the voltage thus, the circuit contains resistance and inductance. Power factorcos φ = 0.6 R √R2+XL 2 = 0.6 ⇒ R 2 + XL 2 = ( R 0.6 ) 2 ⇒ XL 2 = R 2 (0.6) 2 − R 2 ⇒ XL 2 = R 2 × 0.64 0.36 ∴ XL = 0.8 R 0.6 = 4R 3 ...(i) IInd case Now cos φ = 1 (given) Therefore, circuit is purely resistive, ie, it contains only resistance. This is the condition of resonance in which XL = XC ∴ XC = 4R 3 = 4 3 × 110 3 = 440 9 Ω [From Eq. (i)] 1 2πfC = 440 9 Ω C = 9 2×3.14×60×440 = 54 μF 56 (d) Brightness ∝ Pconsumed ∝ 1 R . For bulb, Rac = Rdc , so brightness will be equal in both the cases 57 (a) Geometric length of a magnet is 6/5 times its magnetic length. ∴ Geometric length =6/5×10=12 cm 58 (c) As B0 = μ0 ni,therefore B0 does not depend upon radius (r) of the solenoid. 59 (b) Charging current, I = E R e − t RC Taking log both sides, Log I = log ( E R ) − t RC When R is doubled, slope of curve increases. Also at t = 0, the current will be less. Graph Q represents the best. 60 (a) Vav = 2 π V0 = 2 π × (Vrms × √2) = 2√2 π . Vrms = 2√2 π × 220 = 198 V 61 (b) L = e di/dt = edt di = 5×10−3 (3−2) H = 5 mH 62 (a) This is a parallel circuit, For oscillation, the energy in L and C will be alternately maximum 63 (a) Eddy currents are produced when a metal is kept in a varying magnetic field. 64 (b) This is because, when frequency v is increased, the capacitive reactance XC = 1 2πvC decreases and hence the current through the bulb increases 65 (c) ES = ns nP EP = 1 200 × 240 = 12 V 66 (b) R = P irms 2 = 240 16 = 15Ω; Z = V i = 100 4 = 25Ω Now XL = √Z 2 − R2 = √(25) 2 − (15) 2 = 20Ω ∴ 2πvL = 20 ⇒ L = 20 2π × 50 = 1 5π Hz 67 (a) If the current is wattles then power is zero. Hence phase difference φ = 90° 68 (b) P = 1 2 V0i0 cos φ ⇒ P = PPeak. cos φ ⇒ 1 2 (Ppeak) = Ppeak cos φ ⇒ cos φ 1 2 ⇒ φ = π 3 71 (c) Here, M = 2H, dφ = 4 Wb, dt = 10 s As φ = M i dφ = M di Or di = dφ M = 4 2 = 2 A Also, dφ = M (di) = 2(1) = 2 Wb 72 (b) The average power consumed in an AC circuit is given by P = V0I0 2 cos φ Where φ is phase angle and V0 and I0 the peak value of voltage and current. Given, V0 = 200V,I0 = 2 A,φ = π 3 . P = 200×2 2 cos π 3 = 200×2 2 × 1 2 = 100 W 73 (a) As M = μ0N1N2A l , therefore, M becomes 4 times