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snow_tiger [21]
4 years ago
10

For the above circuit, use superposition to determine the voltage across the R3 resistor due only to the I1 current source. Use

the following values for your calculations: I1 = 5 A V1= 16 V R1=3 Ohms, R2=12 Ohms, R3=3 Ohms. Find V.
Physics
1 answer:
Tems11 [23]4 years ago
7 0

Answer:

idk

Explanation:

idk

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Can someone help me on question 1?
grigory [225]
In question 1, both of your answers are correct, but I don't understand the process you went through in the 'a' part.

R = v/I . That's a correct formula.
But it doesn't help you in this form, because you need to find I
So turn it into a helpful form ... Solve it for I, so it says I=something.

R= v/I

Multiply each side by I : R I = V.

Now divide each side by R: I= V/R .
THERE'S the equation you want.

I = V / R

I = 1.5 / 10 = 0.15 Amp.

That's slightly cleaner, although I don't really understand what you were actually thinking in that part.

But again ... You answered both parts correctly, and your process in b is fine.
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Which substance has a lower coefficient of friction?
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Slip and slide covered with oil
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What is role of gravity in our daily life?
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Please help me with this! Need to get all schoolwork done!!​
sammy [17]

Explanation:

tomatoes and lemons when used together are acid forming.

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3 years ago
A solid non-conducting sphere of radius R carries a charge Q1 distributed uniformly. The sphere is surrounded by a concentric sp
STALIN [3.7K]

Answer:

E = k Q₁ / r²

Explanation:

For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law

           Ф = ∫ E .dA = q_{int} / ε₀

where Ф the electric flow, qint is the charge inside the surface

To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone

           R <r <R_a

for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.

         E A = q_{int} /ε₀

The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field

          q_{int} = Q₁

The surface area is

          A = 4π r²

we substitute

          E 4π r² = Q₁ /ε₀

          E = 1 / 4πε₀ Q₁ / r²

          k = 1/4πε₀

 

          E = k Q₁ / r²

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