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The voltage in a lightbulb is given by the equation [tex]V = IR[/tex]. In this equation, [tex]V[/tex] is the voltage, [tex]I[/tex] is the current, and [tex]R[/tex] is the resistance.

What is the resistance in a lightbulb with a voltage of 12 V and a current of 0.15 A?

A. [tex]0.013 \Omega[/tex]
B. [tex]1.8 \Omega[/tex]
C. [tex]100 \Omega[/tex]
D. [tex]80 \Omega[/tex]

Sagot :

GinnyAnsweridn
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]
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