What is the current in a \( 3.632 \, \Omega \) resistor connected to a battery that has a \( 0.1382 \, \Omega \) internal resistance when the potential drop across the terminals of the battery is \( 6 \, V \)?

The current in a 3.632 ohm resistor connected to a battery with 0.1382 ohms internal resistance and a 6V potential drop across its terminals is approximately 1.592 A.

To find the current in a resistor connected to a battery, we can use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the total resistance (Rtotal). The total resistance is the sum of the external resistance and the battery's internal resistance. In this case, the potential drop across the terminals of the battery is given as 6V, the external resistance is 3.632 ohms, and the internal resistance of the battery is 0.1382 ohms.

First, we calculate the total resistance:

Rtotal = external resistance + internal resistance

= 3.632 ohms + 0.1382 ohms

= 3.7702 ohms.

Next, we use Ohm's Law to find the current:

I = V / Rtotal

= 6V / 3.7702 ohms.

Therefore, the current in the circuit is approximately 1.592 A.