You short-circuit a 6 V battery by connecting a short wire from one end of the battery to the other end. If the current in the short circuit is measured to be 40 A, what is the internal resistance of the battery?

(b) What is the power generated by the battery?

(c) How much energy is dissipated in the internal resistance every second? (Remember that 1 W = 1 J/s)

The internal resistance of the battery, we can use Ohm's Law: Voltage (V) = Current (I) x Resistance (R). In this case, the voltage of the battery is 6 V and the current in the short circuit is 40 A. Since the wire has negligible resistance, we can assume that the current is flowing only through the battery's internal resistance. Using the formula, we find that the internal resistance is 0.15 ohms.

The power generated by the battery can be calculated using the formula:

Power (P) = Current (I) x Voltage (V).

Substituting the values, we get a power of 240 W.

The amount of energy dissipated in the internal resistance every second can be calculated using the formula:

Energy (E) = Power (P) x Time (T).

Since the power is 240 W and the time is 1 second, the energy dissipated is 240 J.

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