Answer :
Final answer:
The power dissipated by a 2 ohm resistor in both the series circuit with a 6V battery and the parallel circuit with a 4V battery is 8 watts. This is calculated by using Ohm's law and the power formula P=I^2*R.
Explanation:
The student is asking about the power dissipation through a 2 ohm resistor in two different circuits. The first circuit has 6 V battery in series with 1 ohm and 2 ohm resistors, and the second circuit has 4 V battery in parallel with 12 ohm and 2 ohm resistors. To compare the power used in the 2 ohm resistor for both circuits, we can use Ohm's law and the power formula P=I^2*R, where I is the current and R is the resistance.
For the first circuit, the total resistance in series is 1 + 2 = 3 ohms. The current from the battery is I = V/R = 6V / 3 ohms = 2A. The power used in the 2 ohm resistor is P = I^2*R = (2A)^2 * 2 ohms = 8W.
In the second circuit, because the resistors are in parallel, each resistor has the full voltage across it. The current through the 2 ohm resistor is I = V/R = 4V / 2 ohms = 2A. The power used by the 2 ohm resistor is P = I^2*R = (2A)^2 * 2 ohms = 8W.
Therefore, the power dissipated by the 2 ohm resistor is the same for both circuits: 8 watts.
