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