The terminal voltage of a real 12.0-volt battery drops to only 11.2 volts when s

Question

The terminal voltage of a real 12.0-volt battery drops to only 11.2 volts when sending current into a load with a resistance R = 1.50 .

a) What current is flowing through the load?

b) What is the internal resistance of the battery?

c) What is the power dissipated to R and r?

5. (15 pts) The te voltage of a real 12.0-volt battery drops to only 112 volts when sending current into a load with a resistance a) What current is flowing through the load? T12 T 14.9, v 1,502 I t, y 6 A b) What is the internal resistance of the battery? r 12 11,2 2, y 6 c) What is the power dissipated to R and r?

Explanation / Answer

b) 11.2=IR

I=11.2/1.5=7.46A

12-Ir =11.2

Ir=0.8

r=0.8/7.46=0.1042 ohm

c) power dissipated to R=I2R = 7.46*7.46*1.5=83.4774 W
power dissipated to r=I2r = 7.46*7.46*0.1042=5.799 W

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.