A 6 volt battery with 1.2 ohms internal resistance is connected across two light bulbs in parallel whose resistance is 12 ohms each. What is the current flow?

To find the current flow in this circuit, we start with the total resistance and apply Ohm's law. Since the light bulbs are in parallel, we first determine the equivalent resistance of the light bulbs.

For two resistors in parallel, the formula for equivalent resistance (R_eq) is:

[ \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} ]

In this case, both resistors (the light bulbs) have a resistance of 12 ohms:

[ \frac{1}{R_{eq}} = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6} ]

Thus,

[ R_{eq} = 6 \text{ ohms} ]

Next, we consider the internal resistance of the battery, which adds to the total resistance in the circuit. The total resistance (R_total) is the sum of the internal resistance of the battery and the equivalent resistance of the bulbs:

[ R_{total} = R_{internal} + R_{eq} ]

[ R_{total} = 1.2 +