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

learn it on the intertant so you can solve

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.