Question

two groups of cells, each group containing 4 cells in series, are connected in parallel. each cell has em 1.50 v and internal resistance of 0.075 q. the external resistance of the circuit is 2.35 0. determine the current in the 2.35 resistance. 

Answer 1

The total e.m.f of all the cells is equal to the sum of the cells.

So the total e.m.f $E= 4 \times1.5 V=6.0 V$

The internal resistance of the cells in series, $R_{internal}=4 \times0.075 = 0.30 \varOmega$

The internal resistance of the two cells in parallel, $R_{internal}=\frac{1}{\frac{1}{R}} =\frac{1}{\frac{1}{0.3}+ \frac{1}{0.3}} =\frac{1}{\frac{2}{0.3}} = \frac{0.3}{2}=0.15\Omega$

The total internal resistance, $R=2.35+0.15=2.5 \Omega$

The current in the 2.35 resistance, $I= \frac{V}{R}= \frac{6}{2.5}=2.4 A$