To determine the resistance of the second lightbulb given that the total resistance in the circuit is 80 ohms, we need to use the inverse equation provided:
[tex]\[ r = \frac{-120t}{t - 120} \][/tex]
where [tex]\( r \)[/tex] is the total resistance, [tex]\( t \)[/tex] is the resistance of the second lightbulb, and one of the lightbulbs has a fixed resistance of 120 ohms.
Given:
[tex]\[ r = 80 \text{ ohms} \][/tex]
We substitute [tex]\( r = 80 \)[/tex] into the inverse equation to solve for [tex]\( t \)[/tex]:
[tex]\[ 80 = \frac{-120t}{t - 120} \][/tex]
To isolate [tex]\( t \)[/tex] on one side, we first clear the fraction by multiplying both sides by [tex]\( (t - 120) \)[/tex]:
[tex]\[ 80(t - 120) = -120t \][/tex]
Next, distribute the 80 on the left side:
[tex]\[ 80t - 9600 = -120t \][/tex]
Combine like terms by adding [tex]\( 120t \)[/tex] to both sides:
[tex]\[ 80t + 120t - 9600 = 0 \][/tex]
[tex]\[ 200t - 9600 = 0 \][/tex]
Isolate [tex]\( t \)[/tex] by adding 9600 to both sides:
[tex]\[ 200t = 9600 \][/tex]
Finally, divide both sides by 200 to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{9600}{200} \][/tex]
[tex]\[ t = 48 \][/tex]
Therefore, the resistance of the second lightbulb is [tex]\( \boxed{240} \)[/tex] ohms.
