Question

Six lead acid type of secondary cells each of emf $2\, V$ 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)

$\varepsilon_{e q}=(2 \times 6) V=12 V$
$r_{ eq }=6(0.015) \Omega=0.09 \Omega$
$i=\left(\frac{\varepsilon}{R+r_{e q}}\right)=\frac{12 V }{(8.5+0.09) \Omega}$
$V_{\text {terminal }}=\varepsilon-i r=12$
$-\left(\frac{12}{8.5+0.09}\right) 0.09=11.9 V$