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A typical small flashlight contains two batteries, each having an emf of1.5V, connected in series with a bulb having resistance17Ω. (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for1.5hwhat is the total energy delivered to the bulb? (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 somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

Short Answer

Expert verified

(a) If the internal resistance of the batteries is negligible, 0.530Wpower is delivered to the bulb.

(b) If the batteries last for5.0h,9540Jtotal energy delivered to the bulb.

(C) The combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value is 7Ω.

Step by step solution

01

Define the ohm’s law, resistance(R)and power .

According to Ohm’s law, the current flowing through the conductor is directly proportional to the voltage across the two points.

V=IR

Where,Iis current in ampere A,Ris resistance in ohmsΩand V is the potential difference volt V

If real source of emf εhas internal energy r,then its terminal potential difference Vabdepends upon current.

So, formula for the terminal potential difference is:

Vab=ε-Ir


The ratio of to for a particular conductor is called its resistance

or

Where, is resistivity , is length in and is area in .

02

Determine the power.

Given that,

R=17Ω

Also given that, two batteries connected in series with same voltage.

Since, the terminal voltage in the circuit is equal to the voltage of the two batteries

Therefore,V=3.0V

The power dissipated in the circuit is:

P=V2R=3217=0.530W

Hence, if the internal resistance of the batteries is negligible, 0.530Wpower is delivered to the bulb

03

Determine the energy.

Given that,t=5.0h

The energy consumed in given time is:

U=Pt=0.5305×3600=9540J

Hence, if the batteries last for 0.5h,9540Jtotal energy delivered to the bulb.

04

Determine the internal resistance.

The internal resistance of battery is:

r=ε-V1

SubstituteV=IRand I=PRin internal resistance formula

r=ε-IRPR=3.0-0.125170.530/217=3-2.1250.125=7Ω

Hence, the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value is 7Ω.

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