Design a circuit which would provide an output voltage Vout = 3V from a voltage source Vin = 10V assuming that?

Question

What is the purpose of a voltage divider circuit? Design a circuit which
would provide an output voltage Vout = 3V from a voltage source Vin =
10V assuming that power consumption of all resistors cannot exceed
1/4W.

I know what a voltage divider is but how do i create this circuit ?

Answer 1

A voltage divider is exactly what you would expect it to do: it divides down the voltage that is input to it to create a lower voltage. It does this with voltage drops across resistors.



Assume your load is 1K ohms and you want 3 volts across it but you only have a 10 volt source.



the equation for voltage division is Vout = Vin * (R2 / (R1 + R2) ) where R2 is the lower resistance and R1 is the upper resistance into R2. In your case R2 would be the 1K or 1,000 ohm load resistance.

http://en.wikipedia.org/wiki/Voltage_divider



Solving the equation above is:

3V = 10V * (1,000 / (R1 + 1,000) )

3V / 10V = 1,000 / (R1 + 1,000)

0.3 * (R1 + 1,000) = 1,000

0.3*R1 + 300 = 1,000

0.3*R1 = 700

R1 = 700 / 0.3

R1 = 7,000 / 3

R1 = 2,333.3333 ohms



The current thru each resistor is 10V / (R1 + R2) = 10V / 3,333.33 ohms =

0.003 Amps or 3 milliAmps



The power dissipated in the 1K resistor is P = I^2 * R = (0.003)^2 * 1,000 = 0.003 Watts or 3 milliWatts



The power dissipated in the 2,333.3333 resistor is P = I^2 * R = (0.003)^2 * 2,333.3333 = 0.021 Watts or 21 milliWatts



Both of which are less than 1/4 Watt or 0.250 Watt or 250 milliWatts.



The ratio of R1 and R2 is the same, so scaling R1 down can be done, as long as R2 is scaled down the same so that the ratio of R1 to R2 is the same.

e.g. R2 is 250 ohms, then R1 is 583.3333 ohms.



The lowest value for R2 as the 1/4 Watt as the load resistor is 36 ohms and R1 as 84 ohms.

Since R1 + R2 = 36 + 84 = 120 ohms

Notice the the ratio is the same, since 84 is 2.3333 times bigger than 36 ohms.



And the current thru each is

I = V / R = 10V / 120 ohms = 0.0833 Amps



so the wattage for 36 ohms is P = I^2 * R = (0.0833A)^2 * 36 ohms = 0.25 Watts,

the maximum wattage for *each* resistor CASE SIZE RATING.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

if you want the maximum power for ALL resistors not to exceed 0.25 Watts,

then the sum of the resistors must not go lower than:

P = V^2 / R

0.25 = (10V)^2 / R

R = 100V^2 / 0.25

R = 400 ohms



R = R1 + R2

R = 2.3333*R2 * R2

R = 3.33333*R2

400 / 3.33333 = R2

R2 = 120 ohms

R1 = 280 ohms {minimum load resistance}

Answer 2

2

Answer 3

Vout = Vin*R1/(R1+R2)

3/10 = R1/(R1+R2)

We know that R1 = 3Ω and R2 = 7Ω would work but that would be far too much power

I²*R = 0.25 or is less than or equal to 1/4We could use 30 and 70 and I = 0.1 = 100ma

0.1² = 0.01

7*0.01 = .07 < 0.25 so the circuit is a 30Ω and a 70V in series with the 10V supply and there are 3V across the 30Ω