A resistor with R = 9.47 ohms resistance is connected to a real (i.e. non-ideal)

Question

A resistor with R = 9.47 ohms resistance is connected to a real (i.e. non-ideal) battery as shown in the figure.

The battery produces an electromotive force of epsilon ? = 19.7 V. When it is connected to the load resistor above, the terminal voltage of the battery drops down to ?Vt(Change in V^t)= 10.8 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?

Explanation / Answer

i = 10.8/r & i = 19.7/(r+9.47) equating these 2, 10.8/r = 19.7/(r+9.47) ? 1) r = 11.49 O 2) i = 10.8/r = 0.9398 amp 3) P = i²*R = 0.9398²*9.47 = 8.3642 W 4) Pi = i²*r = 0.9398²*11.59 = 10.2938 W 5) efficiency = 8.3642/10.2938 = 81.70%

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