Abattery with an emf of 1.50 v has an internal resistance r. when connected to a resistor r, the terminal voltage is 1.40 v and the current is 0.17 a. what is the value of the external resistor r?

R=7.12ohms

Explanation:

Emf (electromotive force) is the source of energy in an electric circuit. It is also the sum of all potential drops across loads in an electric circuit.

Resistance is the opposition to the flow of current in an electric circuit. The presence of a load therefore tends to restrict the current flowing through it.

According to ohm's law, E=IR

E is the emf

I is the current

R is effective resistance (load can be connected in series or parallel)

According to the question, Emf = 1.50V, current(I) =0.17A and terminal voltage V =1.40V

First we need to get the internal resistance (r), from the formula

Terminal voltage V = E -Ir where r is the internal resistance

Substituting

1.40=1.50-0.17r

1.40-1.50=-0.17r

-0.1=-0.17r

r=-0.17/-0.1

r= 1.7ohms

To get R, using E=I(R+r) since the load is in series with internal resistance,we add both to get our effective resistance

1.50=0.17(R+1.7)

1.50=0.17R+0.289

1.50-0.289=0.17R

1.211=0.17R

R=1.211/0.17

R=7.12ohms.

connect the batteries in series and hope