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Norma-Jean [14]
3 years ago
11

The total source voltage in the circuit is 6-3i V. What is the voltage at the middle source

Mathematics
1 answer:
Bezzdna [24]3 years ago
5 0

Answer:

<em>The voltage at the middle source is</em> (2-4\mathbf{i})\ V

Step-by-step explanation:

<u>Voltage Sources in Series</u>

When two or more voltage sources are connected in series, the total voltage is the sum of the individual voltages of each source.

The figure shown has three voltage sources of values:

2 + 6\mathbf{i}

a + b\mathbf{i}

2 - 5\mathbf{i}

The sum of these voltages is:

V_t=4+a+(6+b-5)\mathbf{i}

Operating:

V_t=4+a+(1+b)\mathbf{i}

We know the total voltage is 6-3\mathbf{i}, thus:

4+a+(1+b)\mathbf{i}=6-3\mathbf{i}

Equating the real parts and the imaginary parts independently:

4+a=6

1+b=-3

Solving each equation:

a = 2

b = -4

The voltage at the middle source is (2-4\mathbf{i})\ V

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3 years ago
Compute the surface area of the portion of the sphere with center the origin and radius 4 that lies inside the cylinder x^2+y^2=
Tom [10]

Answer:

16π

Step-by-step explanation:

Given that:

The sphere of the radius = x^2 + y^2 +z^2 = 4^2

z^2 = 16 -x^2 -y^2

z = \sqrt{16-x^2-y^2}

The partial derivatives of Z_x = \dfrac{-2x}{2 \sqrt{16-x^2 -y^2}}

Z_x = \dfrac{-x}{\sqrt{16-x^2 -y^2}}

Similarly;

Z_y = \dfrac{-y}{\sqrt{16-x^2 -y^2}}

∴

dS = \sqrt{1 + Z_x^2 +Z_y^2} \ \ . dA

=\sqrt{1 + \dfrac{x^2}{16-x^2-y^2} + \dfrac{y^2}{16-x^2-y^2} }\ \ .dA

=\sqrt{ \dfrac{16}{16-x^2-y^2}  }\ \ .dA

=\dfrac{4}{\sqrt{ 16-x^2-y^2}  }\ \ .dA

Now; the region R = x² + y² = 12

Let;

x = rcosθ = x; x varies from 0 to 2π

y = rsinθ = y; y varies from 0 to \sqrt{12}

dA = rdrdθ

∴

The surface area S = \int \limits _R \int \ dS

=  \int \limits _R\int  \ \dfrac{4}{\sqrt{ 16-x^2 -y^2} } \ dA

= \int \limits^{2 \pi}_{0} } \int^{\sqrt{12}}_{0} \dfrac{4}{\sqrt{16-r^2}} \  \ rdrd \theta

= 2 \pi \int^{\sqrt{12}}_{0} \ \dfrac{4r}{\sqrt{16-r^2}}\ dr

= 2 \pi \times 4 \Bigg [ \dfrac{\sqrt{16-r^2}}{\dfrac{1}{2}(-2)} \Bigg]^{\sqrt{12}}_{0}

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4 0
3 years ago
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Kamila [148]

Answer:

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Answer:

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