Question

question 6 of 25
the voltage in a lightbulb is given by the equation ( v = ir ). in this equation, ( v ) is
the voltage, ( i ) is the current, and ( r ) is the resistance. what is the current in a
lightbulb with a voltage of ( 35.0 mathrm{~v} ) and a resistance of ( 175 omega ) ?

a. ( 5.00 mathrm{~a} )
b. ( 1.40 mathrm{~a} )
c. ( 6125 mathrm{~a} )
d. ( 0.200 mathrm{~a} )

Explanation:

Step1: Rearrange the formula for current

Given \(V = IR\), we can solve for \(I\) (current) by dividing both sides of the equation by \(R\). So \(I=\frac{V}{R}\).

Step2: Substitute the given values

We are given \(V = 35.0\space V\) and \(R=175\space\Omega\). Substitute these into the formula: \(I=\frac{35.0}{175}\).

Step3: Calculate the value of \(I\)

\(I=\frac{35.0}{175}= 0.200\space A\)

Answer:

D. \(0.200\space A\)