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.

Solved

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

explanation:

considering the bohr theory, the type of connection that requires the transfer the answer conuldn't be right

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