Question

A series battery of six lead accumulators, each of emf 2.0V and internal resistance 0.50$\Omega$ is charged by a 100 V dc supply. The series resistance should be used in the charging circuit in order to limit the current to 8.0A is

• A. 4$\Omega$
• B. 6$\Omega$
• C. 8$\Omega$
• D. 10$\Omega$

Answer 1

Answer: (C)

8$\Omega$

$\Rightarrow (100-12)=I[3+R]$
$\frac{88}{3+R}=8$
$\Rightarrow R=8\Omega$