A battery has an internal resistance of 0.013Ω and an emf of 9.00 V. What is the maximum current that can be drawn from the battery without the terminal voltage dropping below 8.75 V?

19.23 A

Explanation:

Current: This can be defined as the rate of flow of electric charge in a circuit. The S. I unit of current is Ampere (A).

The Expression for current is given as,

Vd = Ir ... Equation 1

Where Vd = Voltage drop, I = current, r = internal resistant.

Make I the subject of the equation

I = Vd/r ... Equation 2

But,

Vd = E-V ... Equation 3

Where E = Emf, V = Terminal Voltage.

Substitute equation 3 into equation 2

I = (E-V) / r ... Equation 4

Given: E = 9.00 V, V = 8.75 V, r = 0.013 Ω.

Substitute into equation 4

I = (9-8.75) / 0.013

I = 0.25/0.013

I = 19.23 A