Question

An ideal battery would produce an extraordinarily large current if "shorted" by connecting the positive and negative terminals with a short wire of very low resistance. Real batteries do not. The current of a real battery is limited by the fact that the battery itself has resistance. What is the resistance of a $9.0\mathrm{V}$ battery that produces a $21A$ current when shorted by a wire of negligible resistance?

Short answer

The resistance of a$9.0V$battery that produces a$21A$current when shorted by a wire of negligible resistance is$429\mathrm{m}\Omega .$

Step-by-step solution

Given Information

Voltage$=9.0V$

Current$=21A$

Explanation

By applying Ohm's law, we can find the resistance,

$r=\frac{U}{I}\to \left(1\right)$

Assuming $r$is the battery's internal resistance (which is always denoted by $r$when compared with external resistances).

Substitute the values in equation (1)

$r=\left(\frac{9V}{21A}\right)$

$=0.429\Omega$

Final Answer

Hence, the resistance is$429m\Omega .$