To determine the resistance in a lightbulb given the voltage and current, we use Ohm's Law, which is stated as:
[tex]\[ V = I \cdot R \][/tex]
where:
- [tex]\( V \)[/tex] is the voltage,
- [tex]\( I \)[/tex] is the current,
- [tex]\( R \)[/tex] is the resistance.
We need to solve for the resistance [tex]\( R \)[/tex]. Rearranging the equation to solve for [tex]\( R \)[/tex], we have:
[tex]\[ R = \frac{V}{I} \][/tex]
Given:
- Voltage [tex]\( V = 12 \)[/tex] volts,
- Current [tex]\( I = 0.15 \)[/tex] amperes.
Now, substituting the given values into the equation for resistance:
[tex]\[ R = \frac{12}{0.15} \][/tex]
Calculating this provides us with:
[tex]\[ R = 80.0 \, \Omega \][/tex]
Therefore, the resistance in the lightbulb is:
D. [tex]\( 80 \, \Omega \)[/tex]
