Question

A combination of secondary cells has an equivalent EMF of 2 V, internal resistance of 400 $\Omega$. Maximum current that can be drawn is.

• A. 0.005 A
• B. 0.0005 A
• C. 0.010 A
• D. 0.010 A

Answer 1

The correct option is A 0.005 A

Ok. So what the question tells you is that the battery comes with a resistance of 400$\Omega$ of its own. Now we want to maximize the current we can draw from this resistor. That can only be a maximum if we connect the least possible resistance across it. That would be 0 $\Omega$. Connecting 0 $\Omega$ across the terminals of the battery is called shorting the terminals of the battery. Now that means we have a circuit like this –

The current will simply be $i=\frac{V}{R}$
$i=\frac{2}{400}=0.005A$