Title 13 Current of Electricity Tutorial Soln.pdf ✅

St. Andrew’s Junior College H2 Physics 13 - 4 (ii) Consider I = = Anvq Anev . Since both filaments are identical, Ane is constant. Hence the current I through the filament is directly proportional to the mean drift velocity v of the electrons in the filament i.e. I  v . The current through Y is twice the current through Z as the potential difference across Y is twice the potential difference across Z (OR the total current flowing through 2 bulbs and 4 bulbs in parallel are equal). Therefore, the mean drift velocity of the electrons in Y is twice that of the electrons in Z. 6(i) (Using maximum power theorem, r = R when P across R is maximum.) Hence, at max power, P = V2 /R 1.13 = (1.502 )/R R = 1.99 Ω Hence, r = 1.99 Ω [1] [1] (ii) Larger p.d. across R means smaller p.d. across r, since e.m.f. is a constant. Smaller power dissipation across r at larger values of V since power = current × potential difference, and current is the same for R and r [1] [1] [1] 7(i) Since current = 0.30 A, p.d. across each lamp = 2.5 V (read off from graph) Hence, p.d. across each connecting wire = (8.7 – 7.5) / 2 = 0.60 V Therefore, resistance of each connecting wire = 0.60 / 0.30 = 2.0 Ω [1] [1] (ii) straight line through origin with gradient of 0.5 [1] [1] (iii) power loss = I 2R = (0.302 )(2.0) = 0.18 W [1] [1] (iv) 1. R = ρL/A 2.0 = (1.7 × 10−8)(L) / (0.40 × 10−6) L = 47 m [1] [1] 2. I = nAvq v = 0.30 / (8.5 × 1028 × 0.40 × 10−6 × 1.6 × 10-19) = 5.5 × 10-5 m s-1 [1] [1]