Janine determines that the total resistance in her circuit is 80 ohms. Using the inverse equation modeling this situation, find the resistance of the second lightbulb.

The resistance of the second lightbulb is ohms.

A. 120
B. 240
C. 300
D. 40

The correct option for the resistance of the second light bulb in ohms (Ω) is option B;

B. 240

Reason why option B is the correct answer

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,

Where;

r = The resistance of the second light bulb

Unknown parameters;

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;

1/t = 1/120 + 1/r

∴

Therefore, by finding the inverse of both sides of the above equation, we get Janine's model as follows;

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

Plugging in t = 80 in t = 120·r/(r + 120), gives;

80 = 120·r/(r + 120)

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

Learn more about series and parallel circuits here;

b. an equilateral triangle inscribed in a circle

step-by-step explanation:

we are given the figure of two circles having a triangle inside.

now, we know that,

"when a figure is lying inside an another figure, then the outside figure is known as circumscribed figure and the figure lying inside is referred to as the inscribed figure".

as we see that, the circles are containing the triangle completely.

thus, the circles are the circumscribed figures and the triangle is the inscribed figure.

hence, we get, 'an equilateral triangle is inscribed in a circle'.

2.2°  

step-by-step explanation:

let θ = the angle of elevation.

tanθ =77/2020

tanθ = 0.038 12

      θ = tan⁻¹0.038 12

      θ = 2.2°