A circuit consists of a battery and a resistor. The resistor dissipates 2W. Now the resistor is replaced with one that has twice the resistance of the original. What power does the new resistor dissipate
The Correct Answer and Explanation is:
To solve this problem, we can use the power formula for resistive circuits: P=V2RP = \frac{V^2}{R}P=RV2
Where:
- PPP is the power dissipated,
- VVV is the voltage across the resistor,
- RRR is the resistance of the resistor.
We are told that the resistor initially dissipates 2W, so we have: P1=2WP_1 = 2WP1=2W
Let the initial resistance be R1R_1R1, and the voltage across the resistor be VVV. Using the power formula: P1=V2R1P_1 = \frac{V^2}{R_1}P1=R1V2
Substituting the given value of power: 2=V2R12 = \frac{V^2}{R_1}2=R1V2
This equation expresses the relationship between the power, voltage, and resistance for the initial resistor.
Now, we replace the resistor with one that has twice the resistance, so the new resistance is: R2=2R1R_2 = 2R_1R2=2R1
The power dissipated by the new resistor, P2P_2P2, is given by: P2=V2R2P_2 = \frac{V^2}{R_2}P2=R2V2
Substituting R2=2R1R_2 = 2R_1R2=2R1 into the equation: P2=V22R1P_2 = \frac{V^2}{2R_1}P2=2R1V2
We already know from the first equation that V2R1=2\frac{V^2}{R_1} = 2R1V2=2, so: P2=12×2=1WP_2 = \frac{1}{2} \times 2 = 1WP2=21×2=1W
Thus, the power dissipated by the new resistor is 1W.
Explanation:
The power dissipated in a resistor depends on the voltage across it and its resistance. In this case, when the resistance is doubled while keeping the voltage constant, the power dissipated is halved. This occurs because the power is inversely proportional to resistance when the voltage remains unchanged. Therefore, the new resistor dissipates half the power of the original one, which is 1W.
