Question

A circuit is arranged as shown. Then the current from $A$ to $B$ is

• A. +500 mA
• B. +250 mA
• C. -250 mA
• D. -500 mA

Answer 1

Answer: (A)

$I=\frac{\sum (E/r)}{1+R\Sigma (1/r)}$
$=\frac{\frac{5}{15}+\frac{10}{10}}{1+10\left[\frac{1}{15}+\frac{1}{10}\right]}=\frac{1}{2}A=500mA$