Question

Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7 V, and 24.6 V. The current drain on the source must be limited to no more than 1 mA. The number of resistors, their values, and their wattage ratings must be specified. A schematic showing the circuit arrangement and resistor placement must be provided

Answer 1

R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor

Step-by-step explanation:

For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)

Therefore we build a circuit with 4 resistors in series, in such a way that

V = i R

let the voltage

1st resistance

V = i R

R₁ = V / i

R₁ = 14.7 / 1 10⁻³

R₁ = 14.7 10³ Ω

power is

P = V i

P = 14.7 1 10⁻³

P = 14.7 10⁻³ W = 0.0147 W

a resistance of ⅛ W is indicated

2nd resistance

R₂ = 8.18 / 1 10⁻³

R₂ = 8.18 10³ Ω

Power

P = 8.18 1 10⁻³

P = 0.00818W

a 1/8 W resistor

3rd resistance

this resistance is calculated in such a way that

V₁ + V₂ + V₃ = 24.6

V₃ = 24.6 - V₁ -V₂

V₃ = 24.6 - 14.7 - 8.18

V₃ = 1.72 V

R₃ = 1.72 / 1 10⁻³

R₃ = 1.72 10³ Ω

power

P = Vi

P = 1.72 10⁻³

P = 0.00172 W

a resistance of ⅛ W

To obtain the voltage of 24.6 we use this three resistors together

4th resistance

The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage

30 = V₁ + V₂ + V₃ + V₄

V₄ = 30 - V₁ -V₂ -V₃

V₄ = 30 -14.7 - 8.18 - 1.72

V₄ = 5.4 V

R₄ = 5.4 / 1 10⁻³

R₄ = 5.4 10³ Ω

Power

P = V i

P = 5.4 10⁻³

P = 0.0054 W

⅛ W resistance

The values ​​of these resistance are commercially

Let's check the consumption of the circuit

R_total = R₁ + R₂ + R₃ + R₄

R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³

R_total = 30 10³

the current circulating in the circuit is

i = V / R_total

i = 30/30 10³

i = 1 10⁻³ A

therefore it is within the order requirement.

for connections see attached diagram