Title 18 AC Tutorial Soln.pdf ✅

St. Andrew’s Junior College H2 Physics 18-3 Solutions to Mastery Questions 1(c) (i) Max emf = NBAω = (500)(5.0 × 10-2 )(2.5 × 10-2 )(2π × 50) = 196 V r.m.s. emf = 196 / (√2) = 140 V (Shown) [2] (ii) max emf across secondary coil = (50 / 2000)(196) = 4.9 V [1] (iii) 1. 1. The supply voltage is less than emf of cell (1.2 V) or less than voltage needed across diodes (1.4 V) less than 2.6 V, the sum of emf and voltage need across diodes [1] [1] 2. 3.75 – 4.35 × 10-4 C (less accurate result) OR 3.90 – 4.20 × 10-4 C (more accurate result) [1] [2] 2(a) Root-mean-square value of an alternating current is defined as the value of the steady direct current which would dissipate heat at the same average heating effect /rate in a given resistance as the alternating current. [1] (b)(i) For e.m.f. to be maximum, sin ( ) 1 t = . Hence, max e.m.f. Eo = NBA(2 f ) = (800)(0.50)(5.010-2  8.010-2 )(2)(240/60) = 40.2 V [1] [1] (ii) Io = Vo R = 40.2 0.6+11.4 = 3.35 A (ecf given) [1] (iii) 3 35 2 2 . o rms I I = = = 2.37 A (ecf given) [1] (iv) Label Pmax = Io 2 R = (3.352 )(11.4) = 128 W {135 W (from IoVo) is wrong, because this gives total power to both coil and external resistor but the question say P is power only to the external resistor} Label period correctly (accept ‘T’ or 0.25 s) Correct shape with 2 cycles drawn. Do not accept cosine2 graph. [1] [1] [1] 128 0.25 0.50