Question

Two batteries with emf of 1.5 V and 1.2V and internal resistances of 0.50 ohms and 0.80 ohms, respectively, are connected in parallel with + terminals together.

a)Determine the current they would deliver to a resistance of 2.0 ohms.

b)Would more current be delivered if only one battery were used? If so, which one? Explain.

Answer 1

Solution since I Internal 2 emf's are in say & = 1.5v en = 0.52 parallel, with ove terminals together ; &q=1.20 Mg = 0.82 resistance I E rez Hence net emf, E= E + E2 el r2 I +1 ri az fr 1.5 t 102 0.5 0.8 = 3+1:5 - I 2 +1.25 + 1 0.5 0.8 = = 1.38 V 4.5 3.25 - and net internal resistance, r = 442 = (0.5) (0,8) 9144 0.5 tous 9 = 0.4 0.307 , 1.3 Hence new circuit is (a) _mu R=2r Hence for this circuit, . Is e simply by ohms Law - Rter I = 1:38 = 1.38 = 0.598 A 2+0.307: 2.307 Slepievieel angwer

the cases, so that you & al (b) Now I analyse for you both I can underest and easily. If & and sel are only there, Ij = &- B (Rted) IL = 1.5 - 105 2+0.5 2.5 Ij = 0.6A tum _ and if & and re are 72 only - R I, = 1.2 = 1.2 2 +0.8 2.8 Ig = 01434 S and I= 0.598A Hence if battery of & with internal resistance & is used only current delivered is a little very little) more than I as I = 0.6A and I = 0.598A is 117.1 (very little) Hope you get the point . Ask, if any doubts is Hope you like my solution & my efforts of