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Zigmanuir [339]
3 years ago
8

I don’t get this it’s hella hard

Engineering
1 answer:
qwelly [4]3 years ago
7 0

Answer:

V₂ = 20 V

Vt = 20 V

V₁ = 20 V

V₃ = 20 V

I₁ = 10 mA

I₃ = 3.33 mA

It = 18.33 mA

Rt = 1090.91 Ω

Pt = 0.367 W

P₁ = 0.2 W

P₂ = 0.1 W

P₃ = 0.067 W

Explanation:

Part of the picture is cut off.  I assume there is a voltage source Vt there?

First, use Ohm's law to find V₂.

V = IR

V₂ = (0.005 A) (4000 Ω)

V₂ = 20 V

R₁ and R₃ are in parallel with R₂ and the voltage source Vt.  That means V₁ = V₂ = V₃ = Vt.

V₁ = 20 V

V₃ = 20 V

Vt = 20 V

Now we can use Ohm's law again to find I₁ and I₃.

V = IR

I = V/R

I₁ = (20 V) / (2000 Ω)

I₁ = 0.01 A = 10 mA

I₃ = (20 V) / (6000 Ω)

I₃ = 0.00333 A = 3.33 mA

The current It passing through Vt is the sum of the currents in each branch.

It = I₁ + I₂ + I₃

It = 10 mA + 5 mA + 3.33 mA

It = 18.33 mA

The total resistance is the resistance of the parallel resistors:

1/Rt = 1/R₁ + 1/R₂ + 1/R₃

1/Rt = 1/2000 + 1/4000 + 1/6000

Rt = 1090.91 Ω

Finally, the power is simply each voltage times the corresponding current.

P = IV

Pt = (0.01833 A) (20 V)

Pt = 0.367 W

P₁ = (0.010 A) (20 V)

P₁ = 0.2 W

P₂ = (0.005 A) (20 V)

P₂ = 0.1 W

P₃ = (0.00333 A) (20 V)

P₃ = 0.067 W

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Answer:

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   }

decision(car_price, available_cash); or decision(available_cash, car_price);

Explanation:

using functions in Javascript:

functions; this refers to dividing codes into reusable parts.

e.g function function_name() {

console.log("How are you?");

}

you can call or invoke this function by using its name followed by parenthesis, like this: function_name(). each time the function is called it will   print out "How are you?".

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Hence, from the question given the two parameters "car_price" and "available_cash" respectively, we write the function with name function_name:

function decision(car_price, available_cash) {

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decision(car_price, available_cash); or decision(available_cash, car_price);

7 0
3 years ago
Consider a 50-mH inductor. a. Express the voltage across the inductor and then evaluate it at t = 0.25 s if iL (t) = 5e −2t + 3t
IrinaK [193]

Answer:

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Explanation:

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V=0.05*d(5e-2t+3te-2t-2)/dt

since there is no power of e, I'll assume the power to be 1

V=0.05*(-2+3e-2)

at t=0.25

V=0.15e-0.2

V=0.208

3 0
3 years ago
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...

6 0
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snow_lady [41]

Answer:

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Explanation:

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Required:

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Final temperature = ∆T = 303-273 = 30

S = 11x10^-6

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Explanation:

7 0
2 years ago
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