A resistor with R = 8.04 Ohm resistance is connected to a real (i.e. non-ideal)

Question

A resistor with R = 8.04 Ohm resistance is connected to a real (i.e. non-ideal) battery as shown in the figure. The battery produces an electromotive force of E = 15.1 V. When it is connected to the load resistor above, the terminal voltage of the battery drops down to delta Vt = 10.5 V. What is the internal resistance r of the battery? What is the electric current flowing in the circuit? What s the power dissipated by the external resistor? How much power is dissipated inside the battery? What s the efficiency of the circuit n percent?

Explanation / Answer

i = E / (R + r)
i = 15.1 / (8.04 +r)
V across battery .
15.1 - (15.1r / (8.04 +r) ) = 10.5
4.6*8.04 + 4.6r = 15.1r
r = 3.52 ohm

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i = 15.1 / (8.04 +3.52) = 1.31 A

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P = i^2 R = 1.31^2 x 8.04 =13.71 Watt

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P = VI = 1.31 x 15.1= 19.78 Watt
.........................
.efficeincy = Pout / Pin = 13.71 / 19.78 = 0.693

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