A parallel circuit has 2 light bulbs that each have a resistance, R1 = 34.2
Ω
, R2 = 66.7
Ω
. What is the total resistance of this circuit?

Question

A parallel circuit has 2 light bulbs that each have a resistance, R1 = 34.2 
Ω
 , R2 = 66.7 
Ω
 . What is the total resistance of this circuit?

BlackTom AI · Accepted Answer

Let me restate the problem: We have a parallel circuit with two resistors, R1 = 34.2 Ω and R2 = 66.7 Ω, and we need the total (equivalent) resistance.

In a parallel circuit, the reciprocals of the resistances add: 1/R_total = 1/R1 + 1/R2.

Compute each reciprocal:
- 1/R1 = 1 / 34.2 ≈ 0.02924 Ω⁻¹
- 1/R2 = 1 / 66.7 ≈ 0.01499 Ω⁻¹

Add them: 1/R_total ≈ 0.02924 + 0.01499 ≈ 0.04423 Ω⁻¹

Now take the reciprocal to find the total resistance: R_total ≈ 1 / 0.04423 ≈ 22.61 Ω.

A quick consistency check: in parallel, the total resistance should be less than the smallest individual resistor (which is 34.2 Ω); 22.61 Ω satisfies this, confirming the result.

Answer consideration: The calculation relies on the parallel resistance formula and standard rounding rules based on the given significant figures (the result 22.61 Ω reflects typical rounding from the intermediate sums to three or four significant figures).

Physics 1 Hon - Bowles - p06/p07 P10.15 Lesson 4 - Exit Ticket

A parallel circuit has 2 light bulbs that each have a resistance, R1 = 34.2 Ω , R2 = 66.7 Ω . What is the total resistance of this circuit?

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