For the circuit below, if the sum of R1 and R2 is constrained to equal 100Ω (R1 + R2 = 100Ω), find the values of R1 and R2 such that I = 0.07 mA, 0.059mA, 0.033mA, and 0.018mA. Assume R1 << R3, effectively making R1 in parallel with 10kΩ approximately equal to R1. Do not solve this problem using loop or nodal analysis. Rather, proceed by simply analyzing the circuit as a voltage divider.

Question

Jamison Munoz

Physics

6 January, 18:21

For the circuit below, if the sum of R1 and R2 is constrained to equal 100Ω (R1 + R2 = 100Ω), find the values of R1 and R2 such that I = 0.07 mA, 0.059mA, 0.033mA, and 0.018mA. Assume R1 << R3, effectively making R1 in parallel with 10kΩ approximately equal to R1. Do not solve this problem using loop or nodal analysis. Rather, proceed by simply analyzing the circuit as a voltage divider.

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Answers (1)

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Mackenzie Stafford · Answer 1

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Explanation:

R1 + R2 = 100Ω

The voltage across R1 = 5 * R1 / (R1 + R2 + 500) = 5*R1/600

[R1 parallel R3 is assumed to be equal to R1 R1<<
So 5*R1/600 = i * 10k

i=0.07 mA

So, R1/120 = 0.07 * 10

⇒R1=120*0.7 = 84 Ohm

i=0.059 mA

So, R1/120 = 0.059*10

R1 = 120*0.59 = 70.8O hm

i=0.033 mA

So, R1/120 = 0.033 * 10

R1=120*0.33 = 39.6 Ohm

i=0.018 mA

So, R1/120 = 0.018*10

R1=120 * 0.18 = 21.6 Ohm