Answer :
Final answer:
In both the series circuit with a 6V battery and the parallel circuit with a 4V battery, the 2 Ω resistor consumes 8 watts of power.
Explanation:
When comparing the power used in a 2 Ω resistor for different circuit setups, you must calculate the current running through it and then use the formula for power, which is P = I2R.
Series Circuit with 6V Battery:
In a series circuit with a 6V battery and 1 Ω and 2 Ω resistors:
The total resistance is Rtotal = 1 Ω + 2 Ω = 3 Ω.
Using Ohm's law (V = IR), the total current I is V/R, so I = 6V / 3 Ω = 2 A.
The power P used by the 2 Ω resistor is P = I2R = (2 A)2 x 2 Ω = 8 W.
Parallel Circuit with 4V Battery:
In a parallel circuit with a 4V battery and 120 Ω and 2 Ω resistors:
The voltage across each resistor in parallel is the same as the source, which is 4 V.
- The current through the 2 Ω resistor is I = V/R = 4V / 2 Ω = 2 A.
- The power P used by the 2 Ω resistor is P = I2R = (2 A)2 x 2 Ω = 8 W.
In both the series and parallel circuits described, the power consumed by the 2 Ω resistor is the same, 8 watts.
