Question

3. Two batteries with emf's of 1.5 V and 1.2 V and internal resistances of 0.50 Ω and 0.80 Ω, respectively, are connected in parallel with terminals together. (a) Determine the current they would deliver to an external resistance of 2.0 Ω.

Answer 1

The equivalent emf of the two cells connected in parallel with internal resistancer; and r, is given as, & & +- - i 12 is the emf of the second Here, & is the emf of the first cell with internal resistance r; and cell with internal resistancer. The equivalent resistance of the circuit, when two resistances are connected in parallel is given by the formula, p='12 +2 The equivalent emf of the circuit is, & E=1 _?i & 13 1 1 - + iܐ ܕܐ Substitute 1.5 V for & , 1.2 V for Ej, 0.50 for r; and 0.80 for ry in the above equation.

1.5 V E = 0.50 1.2 V 0.80 0.50 E=1.38 V 0.80 The equivalent resistance of the circuits is, p= !!2 7i+72 (0.50 0 (0.80 02) = 0.50 +0.800 r = 0.312 Part a) The current in the circuit is given by the formula, 1-RFT Substitute 1.38 V for E, 0.310 for r and 2 for R in the above equation.

= 1.38 V *20+0.310 I =0.6 A Part B) If one battery is used instead of two then, current will be more than 0.6 A. This is because, E<& and E<&. Thus, the current drawn in the case for & and & will be more because I OCE.