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# Six lead acid type of secondary cells each of emf $2.0V$ and internal resistance $0.015\Omega$ are joined in series to provide a supply to a resistance of $8.5 \Omega$ What are the current drawn from the supply and its terminal voltage ?

$\begin{array}{1 1} 10.9\;V \\ 11.9\;V \\ 9.9\;V \\ 12.9\;V \end{array}$

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A)
Solution :
Use $I= \large\frac{nE}{nr+R}$
Given $E= 2V$
$n=6$
$nE = 2 \times 6$
$r= 0.015$
$nr= 2 \times 0.015 =0.09 \Omega$
$R= 8.5 \Omega$
Substituting the values
$I=\large\frac{2 \times 6}{6 \times 0.015 +8.5} =\frac{12}{0.09 +8.5}$
$\quad= \large\frac{12}{8.509}$$=1.4 A$
The terminal voltage of battery $V=IR$
$\qquad= 1.4 \times 8.5 =11.9\;V$
Answer : $11.9\;V$