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Vitek1552 [10]
3 years ago
5

Coulomb’s Law describes:a.Electric force exerted on one charged sphere, by another charged sphere.b.Magnetic field surrounding a

current-carrying conductor.c.Amount of heat generated by high current flow in a resistor.d.Voltage induced in a coil by a constant magnetic field.
Physics
1 answer:
Leni [432]3 years ago
7 0

Answer:

a.Electric force exerted on one charged sphere, by another charged sphere.

Explanation:

Coulomb's Law describes the electric force between two objects. It is given by the following formula:

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

The magnetic field surrounding a current carrying conductor is described by Ampere's Law.

Amount of heat generated by a high current flow in a resistor can be described by P = VI

Voltage induced in a coil by a constant magnetic field is described by Faraday's Law.

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For the circuit shown, R = 75.0 ohms, L= 55.0 mg, and C = 25.0 μC. The power source has 12.0 V arms and a frequency of 60.0 Hz.
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Explanation:

Given that,

Resistance R = 75.0 ohms

Inductance L = 55.0 mH

Capacitance C = 25.0\ \mu C

Voltage V = 12.0 V

Frequency f = 60.0 Hz

We need to calculate the angular frequency

Using formula of angular frequency

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X_{L}=\omega\times L

Put the value into the formula

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X_{L}=20.724\ \Omega

(b). We need to calculate the  value of X_{L}

Using formula of X_{C}

X_{C}=\dfrac{1}{\omega C}

X_{C}=\dfrac{1}{376.8\times25.0\times10^{-6}}

X_{C}=106.16\ \Omega

(c). We need to calculate the value of Z

Using formula of impedance

Z=\sqrt{R^2+(X_{L}-X_{C})^2}

Put the value into the formula

Z=\sqrt{75.0^2+(20.724-106.16)^2}

Z=113.68\ \Omega

(d). We need to calculate the rms current

Firstly we need to calculate the current

Using formula of current

I=\dfrac{V}{R}

Put the value into the formula

I=\dfrac{12.0}{75.0}

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Using formula of rms current

I_{rms}=\dfrac{I_{0}}{\sqrt{2}}

I_{rms}=\dfrac{0.16}{\sqrt{2}}

I_{rms}=0.113\ A

(e). We need to calculate the rms voltage across the resistor

Using formula of rms voltage

V_{rms}=I_{rms}\times R

V_{rms}=0.113\times75.0

V_{rms}=8.475\ V

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Using formula of rms voltage

V_{rms}=I_{rms}\times X_{L}

V_{rms}=0.113\times20.724

V_{rms}=2.342\ V

(g). We need to calculate the rms voltage across the capacitor

Using formula of rms voltage

V_{rms}=I_{rms}\times X_{C}

V_{rms}=0.113\times106.16

V_{rms}=11.99\ V

(h).  We need to calculate the dissipated power by the circuit

Using formula of dissipated power

P=RI^2

Put the value into the formula

P=75.0\times0.113^2

P=0.958\ W

Hence, This is the required solution.

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