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A25.0-Ωbulb is connected across the terminals of a12.0-Vbattery having3.50Ωof internal resistance. What percentage of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb?

Short Answer

Expert verified

12.3%of the power of the battery is dissipated across the internal resistance and hence is not available to the bulb.

Step by step solution

01

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

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

V=IR

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

The ratio ofVto Ifor a particular conductor is called its resistanceR

R=VIorpLA

Where,pis resistivity Ω-m,L, is length in mand Ais area inm2.

The power (P)is the product of potential difference (V)and the current (I).

P=VIorI2RorV2R

02

Determine the percentage of the power of the battery dissipated.

Given that,

R=12Ωε=12Vr=35Ω

The power dissipated by the internal energy is

Pr=I2r

The total power consumed by the internal resistance and the bulb is

Pt=I2rr+R

The ratio between the two power is

PrPt=I2rI2r+R=rr+R=3.53.5+25=12.3%

Hence,12.3% of the power of the battery is dissipated across the internal resistance.

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