Question

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?

Answer 1

The maximum current that can be drawn from a battery with an emf of 9.00 V and internal resistance of 0.013Ω without the terminal voltage dropping below 8.75 V is approximately 19.23 A. This is found using the relationship V = emf - Ir by solving for I.

Step-by-step explanation:

To determine the maximum current that can be drawn from a battery without the terminal voltage dropping below a certain value, we can use Ohm's law and the relationship between emf, current, internal resistance, and terminal voltage. The terminal voltage V is given by the emf minus the product of the current I and the internal resistance r, or V = emf - Ir. Given a battery with an emf of 9.00 V and an internal resistance of 0.013Ω, and requiring that the terminal voltage remains above 8.75 V, we can set up the equation 8.75 V = 9.00 V - I(0.013Ω) to find the maximum current I.

Subtracting the emf from both sides gives -0.25 = -I(0.013Ω), which can be rearranged to find I = 0.25 / 0.013. Solving for I gives a maximum current of approximately 19.23 A. Accordingly, the maximum current that can be drawn from this battery without the terminal voltage dropping below 8.75 V is roughly 19.23 A.

Answer 2

19.23 A

Step-by-step 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