(b) If R 1 = R 2 = 9.9 k?, what is the smallest value of R load that can be used

Question

(b) If R1 = R2 = 9.9 k?, what is the smallest value of Rload that can be used so that Vout drops by less than 11 percent from its unloaded value? (Vout is measured with respect to ground.)

Just need the answer for part b, thanks.

The circuit fragment shown in the figure below is called a voltage divider Ri R2 Rload (a) If Rload is not attached, show that Vout VR2/(R1 + R2). (Do this on paper. Your instructor may ask you to turn in this work.) (b) If R1 = R2 9.9 kQ, what is the smallest value of Rload that can be used so that Vout drops by less than 11 percent from its unloaded value? (Vout is measured with respect to ground.) 1.167977528 Xk

Explanation / Answer

From the circuit diagram, unloaded voltage is

Vout =(R2/R1+R2)*V =[9.9/(9.9+9.9)]*V =0.5V

Now the loaded output voltage is VL

Vout-VL/Vout =11/100

100Vout -100VL =11Vout

VL =0.89Vout =0.89*0.5V =0.445V-----(1)

Now from the circuit diagram here R1=R2=R

VL =[(RLR/RL+R)/(RLR/RL+R)+R]V =[RL/R+2RL]*V----(2)

Now from the equation (1) and (2) we get

RL/R+2RL]*V=0.445V

RL/R+2RL =0.445

RL =0.445R+0.89RL

0.11RL =0.445R ==>RL =(0.445/0.11)R =4.045R =4.045*9.9*1030hm =36.409kohm

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