RI 18 Alternating Currents tutorial solutions

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 1 Tutorial 18 Alternating Currents Suggested Solutions D1 [C] In order to melt identical fuse, power dissipated in the fuse by the d.c. source must be equal to the mean power dissipated in the fuse by the a.c. source: Hence, if a fuse melts when alternating current just exceeds 13 A r.m.s., it will also melt when direct current just exceeds 13 A. D2 In Fig. 2a, mean power ( ) 2 22 0r.m.s.1 0 2 2 VVVP R RR= = = In Fig. 2b, mean power 22 r.m.s.2 0'2 VVPP RR= = = D3 (a) Case (i) and (ii): r.m.s. 10 7.07 A 2 = =I (b) The r.m.s. current of sinusoidal waveforms is independent of frequency or period. (c) Case (iii): Using graphical method, 1. Square the instantaneous current I 2. Mean square current = (total area in one period) / (time for one period) (50 0.01) 0 250.02 ×+= = A2 3. Take square root of the mean square current 5= A Therefore 5 A=Ir.m.s. These steps prove that, for half-wave rectified current with peak value I0, the r.m.s. current is given by 0 2= IIr.m.s. 22 dc 2 dc rms dc ac ac ac PP RR = = = = I I I I I 0 0 100 t / s 0.01 0.02 0.03 I 2 / A2 50

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 2 Alternatively: Case (iv): 22 r.m.s. (15 0.02) (10 0.02) 12.7 A0.04 × +×= =I D4 (a) Total time in 2 periods = 5 cm × 0.5 ms cm-1 = 2.5 ms Period = 1.25 ms (b) Frequency = 1/(1.25 ms) = 800 Hz (c) Peak value of potential difference = (4 ⅔ cm × 2 V cm-1) / 2 = 4.7 V (d) Root-mean-square value of potential difference = 4.7 3.3 V 2 = 152 t / s 0.02 0.04 0.06 I 2 / A2 102 0 0 2 periods 2 × peak value ( ) ( ) ( ) ( ) ∝ ∝ ⇒= ⇒ = 2 r.m.s. 2 r.m.s. 2 r.m.s. r.m.s.2 r.m.s. For full sinusoidal waveform, mean power delivered is ----- (1) For half-wave rectified waveform, half the mean power is "lost", ' ----- (2)2 '(2) 1 '(1) 2 P P I I I I I = = = = 00 r.m.s. 1 1 10 A 5 A222 22 III

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 3 D5 [D] =  = = ×= = = = =×= ss pp s sp p s 3200 15 240 V r.m.s.200 240 2.0 A120 240 2.0 480 W s VN VN NVV N VI R P D6 (a) (i) ss s sp pp p 40 230 9.2 V r.m.s.1000 VN N VVVN N= ⇒= = ×= (ii) peak 9.2 2 13.0 VV = = (b) (i) To provide half-wave rectification such that current only flows in a direction (clockwise) that charges the battery. If the current flows in the opposite direction, it will discharge the battery. (ii) To increase overall resistance of circuit in order to reduce charging current  restricts the initial charging current which may be large if the battery’s initial e.m.f. is low as compared to the p.d. across the secondary coil of the transformer (9.2 V r.m.s.).  less heat is generated within the battery once it is fully charged, maintaining the battery temperature within its safe limits D7 [B] Period is the same. The statements for option A, C and D are correct. D8 [A] Note: Potential difference VXY = VX – VY = potential of X with respect to Y. R X Y Z A B C D Vxy t/s 0.02 0 When , diodes A and C are forward biased and conducts, so When , diodes B and D are forward biased and conducts so R X Y Z B D R X Y Z A C Vxz t/s 0.02 0

RAFFLES INSTITUTION YEAR 5-6 PHYSICS DEPARTMENT 4 C1 In mutual induction, changes of current in one coil (the primary) can cause changes in the flux linkages with another coil (the secondary) and an induced e.m.f. is produced. The flux linkage of the secondary NsΦs depends on the primary current Ip which is creating the flux. ss pNMΦ = I where M is the mutual inductance between the two coils, which may or may not be symmetric. M can be shown to be proportional to the product N pNs of their numbers of turns, and dependent only on the geometry of the two coils, not on the current. From this equation, ps s ddNM dt dt Φ = I Or p s dM dtε =− I The negative sign is a reflection of Lenz’s law. Similarly, for the opposite case in which a changing current I s in the secondary causing a changing flux linkage NpΦp and an e.m.f. εp in the primary, we have s p dM dtε =− I From Fig. 23.1, ( ) -1P Q 2 mV 5 A sdMMdtε = −⇒= −I From Fig. 23.2, ( ) -1Q PP 2 A sdMMdtεε = − ⇒= −I Dividing, P P 2 mV 5 2 (2 mV) 0.80 mV25 εε = ⇒= = C2 1500 15t= −I ( ) ( ) 22 00 r.m.s. 0.02 2 0 0.0232 0 1500 15 0.02 2250000 45000 225 0.02 2250000 45000 22532 0.02 75 8.66 A TT dt t dt T t t dt tt t − = = −+ =  −+= = = ∫∫ ∫ I I