A car battery with negligible internal resistance has a terminal voltage of 12 V. The circuit which starts the car has some resistance $R$. When the battery is connected to that circuit, there is a current of 600 A. If the battery develops an internal resistance of 1 $\Omega$, which of the following is closest to the terminal voltage, with the same circuit connected?

A. 12 V
B. 10 V
C. 6 V
D. 2 V
E. 0 V

The internal resistance of a battery causes a voltage drop in the terminal voltage. If the internal resistance was 1 Ohm for a battery with a 12 V emf and a 600A current, this would result in a negative terminal voltage, meaning there's a likely error in the data. An internal resistance of 0.01 Ohm would yield a terminal voltage closest to 6 V.

Explanation:

When a battery has internal resistance, it means that a small amount of voltage will be lost to overcoming that resistance. This is described by the formula V = E - IR, where V is the terminal voltage, E is the emf of the battery (in this case 12 V), I is the current, and R is the internal resistance.

In your situation, the battery develops an internal resistance of 1 Ohm (W is likely a typo, the unit for resistance is Ohm) and the current is 600 A. Plugging these values into the formula, we get V = 12 V - (600 A * 1 Ohm), which simplifies to V = 12 V – 600 V = -588 V. This is not a realistic value; most likely, it is a problem with the data given in the question as you can't have a negative terminal voltage.

However, assuming the internal resistance was supposed to be smaller (e.g., 0.01 ohms, which is a reasonable value), you'd get V = 12 V - (600 A * 0.01 Ohm) = 6 V, so the closest answer would be option C.

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