Question

A battery with an emf of 12.00 V has an internal resistance r. When connected to a resistor R, the terminal voltage is 11.70 V and the current is 0.20 A.

A) What is the value of the external resistor R?

B) What is the internal resistance r of the battery?

C) What is the energy dissipated in the battery's internal resistance in 3.7 minutes?

D) When a second identical battery is added in series and the external resistor is R = 28 Ohms what is the resulting current?

Answer 1

A)

load resistance is

V = I R _load

R_load = V / I = 11.7 V / 0.2 A = 58.5 ohm

B)

The relation between internal resistance and emf

V = emf - IR_in

R_in = emf - V / I

= 12 V - 11V / 0.20 A = 5 ohm

3)

Energy dissipated in internal resistance is

E = (0.20 A)^2 ( 5 ohm) (3.7 s )

= 0.74 J

4)