Answer :
Sure, let's analyze this situation step-by-step using Ohm's Law, which states that [tex]\( V = I R \)[/tex], where [tex]\( V \)[/tex] is the voltage (potential difference), [tex]\( I \)[/tex] is the current, and [tex]\( R \)[/tex] is the resistance.
1. Initially:
- Let [tex]\( V \)[/tex] be the initial potential difference across the resistor.
- Let [tex]\( I \)[/tex] be the initial current through the resistor.
- According to Ohm's Law:
[tex]\[ V = I R \][/tex]
2. After the potential difference increases:
- The new potential difference is [tex]\( 4V \)[/tex] (it increased by a factor of 4).
- Let [tex]\( I_{\text{new}} \)[/tex] be the new current through the resistor.
- According to Ohm's Law, we now have:
[tex]\[ 4V = I_{\text{new}} \cdot R \][/tex]
3. Finding the new current:
- Rearrange the above equation to solve for [tex]\( I_{\text{new}} \)[/tex]:
[tex]\[ I_{\text{new}} = \frac{4V}{R} \][/tex]
4. Express [tex]\( I_{\text{new}} \)[/tex] in terms of the initial current [tex]\( I \)[/tex]:
- Recall that the initial current [tex]\( I \)[/tex] is:
[tex]\[ I = \frac{V}{R} \][/tex]
- Therefore, the new current can be written as:
[tex]\[ I_{\text{new}} = \frac{4V}{R} = 4 \left( \frac{V}{R} \right) = 4I \][/tex]
This shows that the new current [tex]\( I_{\text{new}} \)[/tex] is four times the initial current [tex]\( I \)[/tex].
Thus, when the potential difference across a resistor increases by a factor of 4, the current through the resistor also increases by a factor of 4.
So, the correct answer is:
B. It increases by a factor of 4.
1. Initially:
- Let [tex]\( V \)[/tex] be the initial potential difference across the resistor.
- Let [tex]\( I \)[/tex] be the initial current through the resistor.
- According to Ohm's Law:
[tex]\[ V = I R \][/tex]
2. After the potential difference increases:
- The new potential difference is [tex]\( 4V \)[/tex] (it increased by a factor of 4).
- Let [tex]\( I_{\text{new}} \)[/tex] be the new current through the resistor.
- According to Ohm's Law, we now have:
[tex]\[ 4V = I_{\text{new}} \cdot R \][/tex]
3. Finding the new current:
- Rearrange the above equation to solve for [tex]\( I_{\text{new}} \)[/tex]:
[tex]\[ I_{\text{new}} = \frac{4V}{R} \][/tex]
4. Express [tex]\( I_{\text{new}} \)[/tex] in terms of the initial current [tex]\( I \)[/tex]:
- Recall that the initial current [tex]\( I \)[/tex] is:
[tex]\[ I = \frac{V}{R} \][/tex]
- Therefore, the new current can be written as:
[tex]\[ I_{\text{new}} = \frac{4V}{R} = 4 \left( \frac{V}{R} \right) = 4I \][/tex]
This shows that the new current [tex]\( I_{\text{new}} \)[/tex] is four times the initial current [tex]\( I \)[/tex].
Thus, when the potential difference across a resistor increases by a factor of 4, the current through the resistor also increases by a factor of 4.
So, the correct answer is:
B. It increases by a factor of 4.
