Answer :
Certainly! Let's go through the problem step-by-step to understand how the power dissipated by a resistor changes when the current through it is cut in half.
The power dissipation in a resistor can be described using the formula:
[tex]\[ P = I^2 \cdot R \][/tex]
where:
- [tex]\( P \)[/tex] is the power dissipated,
- [tex]\( I \)[/tex] is the current through the resistor,
- [tex]\( R \)[/tex] is the resistance of the resistor.
Step 1: Initial Power Dissipation
Let's start with an initial current [tex]\( I_{\text{initial}} \)[/tex]. The initial power dissipation [tex]\( P_{\text{initial}} \)[/tex] can be calculated as:
[tex]\[ P_{\text{initial}} = I_{\text{initial}}^2 \cdot R \][/tex]
Step 2: Final Power Dissipation
Now suppose the current is cut in half. Therefore, the new current [tex]\( I_{\text{final}} \)[/tex] is:
[tex]\[ I_{\text{final}} = \frac{I_{\text{initial}}}{2} \][/tex]
The final power dissipation [tex]\( P_{\text{final}} \)[/tex] can be calculated as:
[tex]\[ P_{\text{final}} = \left( \frac{I_{\text{final}}}{2} \right)^2 \cdot R \][/tex]
Substitute [tex]\( I_{\text{final}} \)[/tex]:
[tex]\[ P_{\text{final}} = \left( \frac{I_{\text{initial}}}{2} \right)^2 \cdot R = \frac{I_{\text{initial}}^2}{4} \cdot R \][/tex]
Step 3: Comparing Initial and Final Power Dissipation
Now, we compare the initial power dissipation with the final power dissipation to determine the factor by which the power changes.
[tex]\[ \frac{P_{\text{initial}}}{P_{\text{final}}} = \frac{I_{\text{initial}}^2 \cdot R}{\frac{I_{\text{initial}}^2}{4} \cdot R} \][/tex]
The resistances [tex]\( R \)[/tex] and currents [tex]\( I_{\text{initial}} \)[/tex] can be canceled out:
[tex]\[ \frac{P_{\text{initial}}}{P_{\text{final}}} = \frac{I_{\text{initial}}^2}{\frac{I_{\text{initial}}^2}{4}} = 4 \][/tex]
Conclusion:
The power dissipation decreases by a factor of 4 when the current through the resistor is cut in half. Therefore, the correct answer is:
C. It decreases by a factor of 4.
The power dissipation in a resistor can be described using the formula:
[tex]\[ P = I^2 \cdot R \][/tex]
where:
- [tex]\( P \)[/tex] is the power dissipated,
- [tex]\( I \)[/tex] is the current through the resistor,
- [tex]\( R \)[/tex] is the resistance of the resistor.
Step 1: Initial Power Dissipation
Let's start with an initial current [tex]\( I_{\text{initial}} \)[/tex]. The initial power dissipation [tex]\( P_{\text{initial}} \)[/tex] can be calculated as:
[tex]\[ P_{\text{initial}} = I_{\text{initial}}^2 \cdot R \][/tex]
Step 2: Final Power Dissipation
Now suppose the current is cut in half. Therefore, the new current [tex]\( I_{\text{final}} \)[/tex] is:
[tex]\[ I_{\text{final}} = \frac{I_{\text{initial}}}{2} \][/tex]
The final power dissipation [tex]\( P_{\text{final}} \)[/tex] can be calculated as:
[tex]\[ P_{\text{final}} = \left( \frac{I_{\text{final}}}{2} \right)^2 \cdot R \][/tex]
Substitute [tex]\( I_{\text{final}} \)[/tex]:
[tex]\[ P_{\text{final}} = \left( \frac{I_{\text{initial}}}{2} \right)^2 \cdot R = \frac{I_{\text{initial}}^2}{4} \cdot R \][/tex]
Step 3: Comparing Initial and Final Power Dissipation
Now, we compare the initial power dissipation with the final power dissipation to determine the factor by which the power changes.
[tex]\[ \frac{P_{\text{initial}}}{P_{\text{final}}} = \frac{I_{\text{initial}}^2 \cdot R}{\frac{I_{\text{initial}}^2}{4} \cdot R} \][/tex]
The resistances [tex]\( R \)[/tex] and currents [tex]\( I_{\text{initial}} \)[/tex] can be canceled out:
[tex]\[ \frac{P_{\text{initial}}}{P_{\text{final}}} = \frac{I_{\text{initial}}^2}{\frac{I_{\text{initial}}^2}{4}} = 4 \][/tex]
Conclusion:
The power dissipation decreases by a factor of 4 when the current through the resistor is cut in half. Therefore, the correct answer is:
C. It decreases by a factor of 4.
