A battery with an emf of 12.0 V shows a terminal voltage of 11.3 V when operatin

Question

A battery with an emf of 12.0 V shows a terminal voltage of 11.3V when operating in a circuit with two lightbulbs rated at 3.0 W (at 12.0 V) which are connected in parallel. Part A What is the battery's internal resistance? A battery with an emf of 12.0 V shows a terminal voltage of 11.3V when operating in a circuit with two lightbulbs rated at 3.0 W (at 12.0 V) which are connected in parallel. A battery with an emf of 12.0 V shows a terminal voltage of 11.3V when operating in a circuit with two lightbulbs rated at 3.0 W (at 12.0 V) which are connected in parallel. A battery with an emf of 12.0 V shows a terminal voltage of 11.3V when operating in a circuit with two lightbulbs rated at 3.0 W (at 12.0 V) which are connected in parallel. Part A What is the battery's internal resistance?

Explanation / Answer

effective power of the bulbs connected in parallel= 3+3=6 WATTS

voltage difference across the bulbs= 11.3 volts

P=V X I

6=(11.3)X I

I= 0.531 amperes .................(1)

let "r" be the internal resistance of the battery

12= 11.3 + (I)(r)

substituting (1) in the above equation, we get

r= 1.32 ohms

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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