Consider the vi characteristic of the battery shown in the f

Consider the v-i characteristic of the battery shown in the figure below. Identify the open circuit voltage. Identify the short-circuit current. Find an expression for the internal series resistance of the battery and calculate a value for the resistance. If the battery is connected to a 100 Ohm load, calculate the load voltage. Provide a general expression and calculate the answer. How much heat is dissipated by the resistor? How about within the battery?

Solution

Ans)

a) Open circuit Voltage when current =0 A i.e V=1.5 V from figure

Voc=1.5 V

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b) short circuit current when voltage =0 V i.e I=150 mA

Isc=150 mA

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c) Internal resistance is the Ri=Voc/Isc=1.5/150m=10 ohms

Ri=10 ohms

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d) Voltage across the load using voltage division is VL=Voc*RL/(Ri+RL) where RL= load Resistance

VL=1.5*100/(10+100)=1.364 V

VL=1.364 V

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e)

IL=VL/RL=1.364/100=13.64 mA load current

Power dissipated in RL=IL^2*RL=13.64m^2*100=0.0186 W

Power dissipated in Ri=IL^2*Ri=13.64m^2*10=0.00186 W power within battery

And power dissipated is directly proportional to heat dissipated