A battery cable terminal with 0.01 ohms will experience what voltage drop with one amp of current?

To determine the voltage drop across a battery cable terminal with a resistance of 0.01 ohms when one amp of current flows through it, you can use Ohm's Law, which states that voltage (V) is equal to current (I) multiplied by resistance (R).

In this case, when the current is 1 amp and the resistance is 0.01 ohms, the calculation would be:

V = I × R

V = 1 A × 0.01 Ω

V = 0.01 volts

This calculation confirms that with one amp of current flowing through a terminal that has a resistance of 0.01 ohms, the resulting voltage drop is indeed 0.01 volts, which can also be expressed as one-tenth of a volt. Therefore, the correct answer reflects this calculation, leading to one-tenth of a volt being the resulting voltage drop.

The other options do not correspond to the resistance and current provided in the question. One volt would significantly exceed the calculated drop, while zero volts would imply no resistance or current. Two volts also does not align with the calculated outcome based on Ohm's Law.