Let ?V b = 3.7 V. I) If R1 = 1 k? (1000 Ohms), what should R2 be so that ?VL = ?

Question

Let ?Vb = 3.7 V.

I) If R1 = 1 k? (1000 Ohms), what should R2 be so that ?VL = ?VOut = 2.5 V when RL is infinite (no load resistance)?

II) Using your result from (I), what is the smallest value of RL such that ?VL is within 10% of ?VOut = 2.5 V. In other words, what is the smallest value of RL such that ?VL > 2.25 V?

III) A problem with this circuit is that power is wasted in the resistances R1 and R2, when you want the power supplied to RL. Using the values you found in (I) and (II), determine the electrical power converted to other forms of energy in R1, R2, and RL.

Explanation / Answer

1) Since this is voltage divider circuit so

Vout = R2*Vin/ (R1+R2) => 2.5*(1000+R2) = R2*3.7     => R2*1.2 = 2.5*1000

=> R2 = 2.5*1000/1.2 = 2.08 K Ohms Answer

2) This time we will solve for 2.25V but instead of R2 we will use Req = R2 || RL

So Vout = Req*Vin/ (R1+Req) => 2.25*(1000+Req) = Req*3.7     => Req*1.45 = 2.25*1000

=> Req = 2.25*1000/1.45 = 1.55 KOhms

RL = 1/(1/Req - 1/R2) =1/(1/1.55 - 1/2.08) = 6.083 K ohms Answer

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