Question

Two cells, each of E.M.F. 1.5 V and internal resistance 2Ω are connected in parallel. The battery of cells is connected to an external source to an external resistance of 3 ohm. Calculate:

(i) The total resistance in the circuit

(ii) Current flowing in the external circuit

(iii) The voltage drop in each cell

(iv) The terminal potential difference

Answer 1

Given,
EMF of battery, $\epsilon =1.5V$
Internal resistance, r = 2 $\Omega$
External resistance, R = 3 $\Omega$

(a) Total resistance of circuit, $R_t=\frac{r}{2}+R=\frac{2}{2}+3=4\Omega$
(b) Current flowing through circuit is given by
$i=\frac{\epsilon }{R_t}=\frac{1.5}{4}=0.375A$
(c) Current through the each batter will be $i_1=\raisebox{1ex}{$i$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
Hence, voltage drop in each cell is given by
$V=\epsilon -i_{1}r=1.5-\frac{0.375}{2}\times 2=1.125V$
(d) Terminal potential will be equal to the potential drop across the external resistance R
$V_t=iR=0.375\times 3=1.125V$