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stich3 [128]
3 years ago
8

A 100 Ω resistor is connected in series with a 47 µF capacitor and a source whose maximum voltage is 5 V, operating at 100.0 Hz.

Find the following. • The capacitive reactance of the circuit • The impedance of the circuit • The maximum current in the circuit • The phase angle
Physics
1 answer:
Pani-rosa [81]3 years ago
5 0

Answer:

X_c=-33.86275385\Omega

|Z|=105.5778675\Omega

I=0.04735841062A

\phi=20.78612878\°

Explanation:

The electrical reactance is defined as:

X_c=-\frac{1}{2\pi fC}

Where:

f=Frequency\\C=Capacitance

So, replacing the data provided by the problem:

X_c=\frac{1}{2\pi *100*(47*10^{-6} )} =-33.86275385\Omega

Now, the impedance can be calculated as:

Z=R+jX_c

Where:

R=Resistance\\X_c= Capacitive\hspace{3}reactance

Replacing the data:

Z=100-j33.86275385

In order to find the magnitude of the impedance we can use the next equation:

|Z|=\sqrt{(R^2)+(X_c^2)}=\sqrt{(100)^2+(-33.86275385)^2} =105.5778675\Omega

We can use Ohm's law to find the current:

V=I*Z\\I=\frac{V}{Z}

Therefore the current is:

I=\frac{5}{100-j33.86275385}=0.04485638113+0.01518960593j

And its magnitude is:

|I|=\sqrt{(0.04485638113)^2+(0.01518960593)^2} =0.04735841062\Omega

Finally the phase angle of the current is given by:

\phi=arctan(\frac{0.01518960593}{0.04485638113})=20.78612878\°

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Answer:

C = 1.77 \times 10^{-4} F

Explanation:

As we know that in AC circuit we have

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here we have

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so we will have

x_c = \frac{59}{5.05}

x_c = 11.68 ohm

also we know that

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<h2>Answer: 56.718 min</h2>

Explanation:

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In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

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Where;

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If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

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