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A resistor with R = 8.09 ohm resistance is connected to a real (i.e. non-ideal) battery as shown in the figure.
The battery produces an electromotive force of ? = 16.9 V. When it is connected to the load resistor above, the terminal voltage of the battery drops down to ?Vt = 12.7 V.

A.) What is the internal resistance r of the battery?
B.) What is the electric current flowing in the circuit?
C.) What is the power dissipated by the external resistor?
D.) How much power is dissipated inside the battery?
E.) What is the efficiency of the circuit in percent?

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a) If Voc=16.9 and Vload=12.7 with a load of 8.09 ohms, the current is 12.7/8.09=1.56 amp. This means the battery resistance is (16.9-12.7)/1.56=2.69 ohms

b) 1.56 amp

c) P=V*I=12.7*1.56=19.8 watts

d)1.56*1.56*2.69=6.54watt

e) eff=Pout/Pin=12.7/16.9=75%.
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