A typical small flashlight contains two batteries, each having emf of 1.50 V connected in series with a bulb having a resistance of 14.0. Part A If th

illusiia

illusiia

Answered question

2021-01-15

A typical small flashlight contains two batteries, each having emf of 1.50 V connected in series with a bulb having a resistance of 14.0.
Part A
If the internal resistance of the battery is negligible, what power is delivered to the bulb?
Part B
If the batteries last for a time of 5.70 h, what is the total energy delivered to the bulb?
Part C
The resistance of real batteries increases as they run down.If the initial internal resistance is negligible,what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change some what when the current through the filament changes,because this changes the temperature of the filament and hence there sistivity of the filament wire.)

Answer & Explanation

Aamina Herring

Aamina Herring

Skilled2021-01-16Added 85 answers

Part(A) Power is
P=V2R=(1.5)214=0.160714W
Part(B) Energy E=Power in kilowatts*time
Energy is
E=0.160711000kW×5.7hr=0.000916kwh
Part(C) When power is reduced to half
P2=V2R+r
This step can also be seen like this P=VI=V(VI)=1.5(1.5R+r)
R+r=VV2P=21.51.50.16071=28.00
14+r=28.00
Internal resistance of battery is r=14 ohms

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