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Sagot :
Sure, let's find the current in a 30.0 Ω resistor. To determine the current, we use Ohm's Law, which states that
[tex]\[ V = IR \][/tex]
where:
- [tex]\( V \)[/tex] is the potential difference (voltage) across the resistor,
- [tex]\( I \)[/tex] is the current through the resistor,
- [tex]\( R \)[/tex] is the resistance.
We can rearrange this equation to solve for the current:
[tex]\[ I = \frac{V}{R} \][/tex]
Given:
- Resistance, [tex]\( R = 30.0 \, \Omega \)[/tex]
- Potential difference, [tex]\( V \)[/tex]
To find the current, we need the potential difference, [tex]\( V \)[/tex], across the resistor. Once the voltage is known, we substitute the values into the equation.
Assuming we have a commonly used voltage value of [tex]\( V = 120V \)[/tex], the calculation would be:
[tex]\[ I = \frac{120V}{30.0 \Omega} \][/tex]
[tex]\[ I = 4.00A \][/tex]
So, the current in the 30.0 Ω resistor is [tex]\( 4.00A \)[/tex].
Therefore, the correct answer is:
B. [tex]\( 4.00 \, A \)[/tex]
[tex]\[ V = IR \][/tex]
where:
- [tex]\( V \)[/tex] is the potential difference (voltage) across the resistor,
- [tex]\( I \)[/tex] is the current through the resistor,
- [tex]\( R \)[/tex] is the resistance.
We can rearrange this equation to solve for the current:
[tex]\[ I = \frac{V}{R} \][/tex]
Given:
- Resistance, [tex]\( R = 30.0 \, \Omega \)[/tex]
- Potential difference, [tex]\( V \)[/tex]
To find the current, we need the potential difference, [tex]\( V \)[/tex], across the resistor. Once the voltage is known, we substitute the values into the equation.
Assuming we have a commonly used voltage value of [tex]\( V = 120V \)[/tex], the calculation would be:
[tex]\[ I = \frac{120V}{30.0 \Omega} \][/tex]
[tex]\[ I = 4.00A \][/tex]
So, the current in the 30.0 Ω resistor is [tex]\( 4.00A \)[/tex].
Therefore, the correct answer is:
B. [tex]\( 4.00 \, A \)[/tex]
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