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;

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