Simple Resistive Circuits The Voltage Divider 80117 Express your answer in V to

Question

Simple Resistive Circuits The Voltage Divider 80117 Express your answer in V to three signiticant tigures. View Available Hint(s) -5.81v Previous Answers Correct Adnga load resistance to he vollage divider ciuit causes the actual output voltage to drop bclow the desig value The smaller the load rcsistance, thc larger thc drop Part C The circuil designer wants to change the values of Ri and R so that the design cutput vollage-6 V is achieved when the load resistance is R-200 k2 ralher than at no-load. The actual outout vollage must not drop below 5.4 V when R 100 k2 What is the smallest resistor value that can be used for R? Express your answer in k to three aignificant figures. View Available Hint(s) R1.00 Submit x Incorrect; Try Again; 4 attempts remaining

Explanation / Answer

the voltage divider

a. given vs = 18 V

R1 = 10 k ohms

vo = 6 V

hence

vo = R2/(R1 + R2) vs

1/3 = R2/(10,000 + R2)

R2 + 10000 = 3R2

R2 = 5,000 ohm = 5 k ohms

c. Vo = 6 V

RL = 200 k ohms

rather than Rl = 0

Vo > 5.4 V when Rl = 100 k ohms

smallest value for Rl is

vo = R2*vs/(R1(1 + R2/Rl) + R2)

6 = R2*18/(R1(1 + R2/200,000) + R2)

R1 + R1*R2/200,000 + R2 = 3R2

R1 + R1*R2/200,000 - 2R2 = 0

R2 = R1/(R1/200,000 - 2)

also.

for smallest R1

5.4 = R2*10/(R1(1 + R2/100.000) + R2)

0.54R1 + 0.54R1R2/100,000 -0.46 R2 = 0

R2 = 0.54R1/(0.54R1/100,000 - 0.46)

0.54R1/(0.54R1/100,000 - 0.46) = R1/(R1/200,000 - 2)

(R1/100,000 - 0.46/0.54) = (R1/200,000 - 2)

R1 = 229.629 k ohms

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