A battery with an emf of 24.0 V is connected to a resistive load. If the terminal voltage of the battery is 16.1 V and the current through the load is 3.90 A, what is the internal resistance of the battery?

2.03 Ω

Explanation:

EMF: This can be defined as the potential difference of a cell when it is not delivering any current. The S.I unit of Emf is Volt.

The formula of emf is given as,

E = I(R+r) Equation 1

Where E = Emf, I = current, R = External resistance, r = internal resistance.

Make r the subject of the equation

r = (E-IR)/I Equation 2

Note: From ohm's law, V = IR.

r = (E-V)/I Equation 3

Where V = Terminal voltage

Given: E = 24 V, I = 3.9 A, V = 16.1 V.

Substitute into equation 3

r = (24-16.1)/3.9

r = 7.9/3.9

r = 2.03 Ω

i wanna say it’s between b and d, leaning more towards b.