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 0.120 A. The value of the external resistor is 97.5 ohms and the value of the internal resistor is 2.50 ohms. What is the energy dissipated in the battery's internal resistance in 4.20 minutes?

Answer 1

emf = 12V

Vt = 11.7V and I=0.12A

P = I^2*Rin = (.12)^2 * 2.5 = .036W

E= P*t = (.036)*4.20*60 =9.072J

Answer 2

E*4/(4 + r) = 12.32

E*6/(6 + r) = 13

4*E = 4*12.32 + 12.32*r = 49.28 + 12.32*r [1]

6*E = 78 + 13*r [2]

Multiply [1] by 3, and [2] by 2

12*E = 147.84 + 36.96*r

12*E = 156 + 26*r

147.84 + 36.96*r = 156 + 26*r

10.96*r = 156 - 147.84 = 8.16

R = 0.7445...ohms

E = (1/6)*(78 + 13*r) from [2]

E = 14.6131...volts

Answer 3

P=(v^2/r) *t=.3^2/2.5*4.2*60= 9.072 joules