7) a, b and c please! The internal resistance of a battery whose emf is 28.8 V v

Question

7) a, b and c please!

The internal resistance of a battery whose emf is 28.8 V varies with the current according to the equation r =, where a=0.170 ohm and b=0.020 ohm/A. What is the terminal voltage of the battery when I = 2.0 A? What is the power dissipated in the battery when I = 2.0 A? What is the terminal voltage in the battery when I = 5.0 A? To avoid electrical fires, wirings must be sufficiently thick. What should be the diameter of a Cu wire if it is to carry a maximum current of 9.0 A and produce no more than 2.0 W of heat per meter of length?

Explanation / Answer

Terminal voltage of the battery is the difference between the emf and the voltage developed across the battery due to the internal resistance.

V = emf - Ir

here, r = a + bI

a = 0.17 ohm and b = 0.02 ohm/A

for I = 2A: r = 0.17 + 0.02 (2) = 0.21 ohms

therefore V = 28.8 - 2(0.21) = 28.38 Volts

Power dissipated in the battery = P = I2r = (2)2(0.21) = 0.84 W.

when I = 5 A, r = 0.17 + 0.02(5) = 0.27 ohms

Therefore, V = 28.8 - 5(0.27) = 27.45 Volts.

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