A flashlight has two flashlight batteries, a bulb, power switch, envelope and a

Question

A flashlight has two flashlight batteries, a bulb, power switch, envelope and a reflector. The voltage of the batteries is 1.65V where they are unloaded. The bulb is marked with 2.8V and the power 1.2W (when the voltage across the lamp is 2.8V). When the batteries are in the flashlight and the light is on, it turns out that the voltage is 2.8V above the lamp.
a) Calculate the internal resistance of the batteries. [ohm per battery]
b) Calculate power output in a battery. [W per battery]
c) Calculate the current in the circuit. [A]

Answer with two decimals for them all.

Explanation / Answer

Given

batteries of 1.65 v each , and bulb is marked as 2.8 v with power P = 1.2 W

the

2e- i*2r = 2.8 ----------(1)

for power P = V*i

i = P/V

i = 1.2/2.8 A = 0.429 A

substituting in (1)

2*1.65 - 0.429*2*r = 2.8

r = 0.583 ohm

a) internal internall resistance of battery is r = 0.583 ohm

b) power output in the battery is P = V*i = 1.65*0.429 W = 0.70785 W

c) current in the circuit i = 0.429 A

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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