Answer :
To determine how the current changes when the potential difference across a resistor increases by a factor of 4, we can analyze the situation using Ohm's law, which states [tex]\( V = I \cdot R \)[/tex].
Let's start by setting up the initial conditions and the subsequent change:
1. Initial setup:
- Let the initial voltage be [tex]\( V_{\text{initial}} \)[/tex].
- Let the initial current be [tex]\( I_{\text{initial}} \)[/tex].
- The resistance [tex]\( R \)[/tex] is calculated using Ohm's Law: [tex]\( R = \frac{V_{\text{initial}}}{I_{\text{initial}}} \)[/tex].
2. Change in voltage:
- The problem states that the potential difference (voltage) increases by a factor of 4.
- Therefore, the new voltage [tex]\( V_{\text{new}} \)[/tex] can be expressed as: [tex]\( V_{\text{new}} = 4 \cdot V_{\text{initial}} \)[/tex].
3. Determining the new current:
- Using Ohm's law again with the new voltage, we can solve for the new current [tex]\( I_{\text{new}} \)[/tex].
- Substituting [tex]\( V_{\text{new}} \)[/tex] and the resistance [tex]\( R \)[/tex], we get:
[tex]\[ V_{\text{new}} = I_{\text{new}} \cdot R \][/tex]
- Plugging in [tex]\( V_{\text{new}} = 4 \cdot V_{\text{initial}} \)[/tex] and [tex]\( R = \frac{V_{\text{initial}}}{I_{\text{initial}}} \)[/tex]:
[tex]\[ 4 \cdot V_{\text{initial}} = I_{\text{new}} \cdot \frac{V_{\text{initial}}}{I_{\text{initial}}} \][/tex]
- Simplifying the equation:
[tex]\[ 4 \cdot V_{\text{initial}} = I_{\text{new}} \cdot \frac{V_{\text{initial}}}{I_{\text{initial}}} \][/tex]
[tex]\[ I_{\text{new}} = 4 \cdot I_{\text{initial}} \][/tex]
4. Conclusion:
- The new current is [tex]\( 4 \)[/tex] times the initial current.
Therefore, the current increases by a factor of 4 when the potential difference increases by a factor of 4. This corresponds to option B.
Final Answer: B. It increases by a factor of 4.
Let's start by setting up the initial conditions and the subsequent change:
1. Initial setup:
- Let the initial voltage be [tex]\( V_{\text{initial}} \)[/tex].
- Let the initial current be [tex]\( I_{\text{initial}} \)[/tex].
- The resistance [tex]\( R \)[/tex] is calculated using Ohm's Law: [tex]\( R = \frac{V_{\text{initial}}}{I_{\text{initial}}} \)[/tex].
2. Change in voltage:
- The problem states that the potential difference (voltage) increases by a factor of 4.
- Therefore, the new voltage [tex]\( V_{\text{new}} \)[/tex] can be expressed as: [tex]\( V_{\text{new}} = 4 \cdot V_{\text{initial}} \)[/tex].
3. Determining the new current:
- Using Ohm's law again with the new voltage, we can solve for the new current [tex]\( I_{\text{new}} \)[/tex].
- Substituting [tex]\( V_{\text{new}} \)[/tex] and the resistance [tex]\( R \)[/tex], we get:
[tex]\[ V_{\text{new}} = I_{\text{new}} \cdot R \][/tex]
- Plugging in [tex]\( V_{\text{new}} = 4 \cdot V_{\text{initial}} \)[/tex] and [tex]\( R = \frac{V_{\text{initial}}}{I_{\text{initial}}} \)[/tex]:
[tex]\[ 4 \cdot V_{\text{initial}} = I_{\text{new}} \cdot \frac{V_{\text{initial}}}{I_{\text{initial}}} \][/tex]
- Simplifying the equation:
[tex]\[ 4 \cdot V_{\text{initial}} = I_{\text{new}} \cdot \frac{V_{\text{initial}}}{I_{\text{initial}}} \][/tex]
[tex]\[ I_{\text{new}} = 4 \cdot I_{\text{initial}} \][/tex]
4. Conclusion:
- The new current is [tex]\( 4 \)[/tex] times the initial current.
Therefore, the current increases by a factor of 4 when the potential difference increases by a factor of 4. This corresponds to option B.
Final Answer: B. It increases by a factor of 4.
