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
1 answer:
Answer:
R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω , R₃ = 1.72 10³ Ω , R₄ = 5.4 10³ Ω 1/8 W resistor
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
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Answer:
t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number
Explanation:
To solve the problem/ we first write the differential equation governing the motion. So,
with m = 1 slug and k = 9 lb/ft, the equation becomes
The characteristic equation is
D² + 9 = 0
D = ±√-9 = ±3i
The general solution of the above equation is thus
x(t) = c₁cos3t + c₂sin3t
Now, our initial conditions are
x(0) = -1 ft and x'(0) = -√3 ft/s
differentiating x(t), we have
x'(t) = -3c₁sin3t + 3c₂cos3t
So,
x(0) = c₁cos(3 × 0) + c₂sin(3 × 0)
x(0) = c₁cos(0) + c₂sin(0)
x(0) = c₁ × (1) + c₂ × 0
x(0) = c₁ + 0
x(0) = c₁ = -1
Also,
x'(0) = -3c₁sin(3 × 0) + 3c₂cos(3 × 0)
x'(0) = -3c₁sin(0) + 3c₂cos(0)
x'(0) = -3c₁ × 0 + 3c₂ × 1
x'(0) = 0 + 3c₂
x'(0) = 3c₂ = -√3
c₂ = -√3/3
So,
x(t) = -cos3t - (√3/3)sin3t
Now, we convert x(t) into the form x(t) = Asin(ωt + Φ)
where A = √c₁² + c₂² = √[(-1)² + (-√3/3)²] = √(1 + 1/3) = √4/3 = 2/√3 = 2√3/3 and Ф = tan⁻¹(c₁/c₂) = tan⁻¹(-1/-√3/3) = tan⁻¹(3/√3) = tan⁻¹(√3) = π/3.
Since tanФ > 0, Ф is in the third quadrant. So, Ф = π/3 + π = 4π/3
x(t) = (2√3/3)sin(3t + 4π/3)
So, the velocity v(t) = x'(t) = (2√3)cos(3t + 4π/3)
We now find the times when v(t) = 3 ft/s
So (2√3)cos(3t + 4π/3) = 3
cos(3t + 4π/3) = 3/2√3
cos(3t + 4π/3) = √3/2
(3t + 4π/3) = cos⁻¹(√3/2)
3t + 4π/3 = ±π/6 + 2kπ where k is an integer
3t = ±π/6 + 2kπ - 4π/3
t = ±π/18 + 2kπ/3 - 4π/9
t = π/18 + 2kπ/3 - 4π/9 or -π/18 + 2kπ/3 - 4π/9
t = π/18 - 4π/9 + 2kπ/3 or -π/18 - 4π/9 + 2kπ/3
t = -7π/18 + 2kπ/3 or -π/2 + 2kπ/3
Since t is not less than 0, the values of k ≤ 0 are not included
So when k = 1,
t = 5π/18 and π/6. So,
t = 5π/18 + 2nπ/3 or π/6 + 2nπ/3 where n is a natural number