Respuesta :
The sum of resistors arranged in parallel is the inverse of the sum of the inverses of the magnitudes of the individual resistances
The correct option for the resistance of the second light bulb in ohms (Ω) is option B;
B. 240
The reason why option B is the correct answer is s follows:
Known parameters:
Based on a online search, the question appears to have some parts missing which can be as follows;
The resistance of the first light bulb = 120 Ω
Janine's model of the total resistance of the circuit, [tex]t = \mathbf {\dfrac{120 \cdot r }{r + 120} }[/tex]
Where;
r = The resistance of the second light bulb
The unknown parameter:
Resistance of the second light bulb
Method:
Find r using Janine's model of the total resistance, which is the equation of total resistances in parallel arrangement
The inverse relationship modelling the sum, t, of resistances, r, and 120, arranged in parallel, presented as follows;
[tex]\mathbf {\dfrac{1}{t} } =\dfrac{1}{120} + \dfrac{1}{r}[/tex]
∴ [tex]\mathbf {\dfrac{1}{t}} = \dfrac{r + 120 }{120 \cdot r}[/tex]
Therefore, by finding the inverse of both sides of the above equation, we get Janine's model as follows;
[tex]t = \mathbf {\dfrac{120 \cdot r }{r + 120} }[/tex]
The above equation is the inverse equation modelling the total resistance of the parallel arrangement of the resistances in the lightbulb
The question details include:
The total resistance in her circuit, t = 80 Ω
Solution:
Plugging in t = 80 in [tex]t = \mathbf {\dfrac{120 \cdot r }{r + 120} }[/tex], gives;
[tex]80 = \mathbf {\dfrac{120 \cdot r }{r + 120} }[/tex]
Therefore, we get;
80·(r + 120) = 120·r
80·r + 80 × 120 = 120·r
∴ 120·r - 80·r = 80 × 120 = 9,600
120·r - 80·r = 40·r
∴ 40·r = 9,600
r = 9,600/40 = 240
The resistance of the second light bulb, r = 240 Ω
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Answer:240
Step-by-step explanation:
i watched the walk through
