Question

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.

Answer 1

Solved

Step-by-step 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<<<R3]

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