Question

Six lead acid type of secondary cells each of emf $2V$ and internal resistance $0.015\Omega$ are joined in series to provide a supply to a $8.5\Omega$ resistance. The terminal voltage of the battery is

• A. 16.2 V
• B. 5.8 V
• C. 22.5 V
• D. 11.9 V

Answer 1

Answer: (D)

${\epsilon }_{eq}=(2\times 6)V=12V$
$r_{eq}=6(0.015)\Omega =0.09\Omega$
$i=\left(\frac{\epsilon }{R+r_{eq}}\right)=\frac{12V}{(8.5+0.09)\Omega }$
$V_{\text{terminal }}=\epsilon -ir=12$
$-\left(\frac{12}{8.5+0.09}\right)0.09=11.9V$