Question

Six lead-acid type of secondary cells each of emf $2.0V$ and internal resistance $0.15\Omega$ joined in series to provide a supply to a resistance of $15\Omega$. What are the current drawn from the supply and its terminal voltage?

• A. $1.4A,11.9V$
• B. $0.75A,11.25V$
• C. $1.4A,7.5V$
• D. $0.5A,11.9V$

Answer 1

The correct option is B $0.75A,11.25V$

$I=\frac{nE}{nr+R}E=2vr=0.15\Omega R=15\Omega n=6\therefore I=\frac{6\times 2}{(6\times 0.15)+15}=\frac{12}{15.9}=0.75AV=IR=0.75\times 15=11.25V$