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tatuchka [14]
3 years ago
7

A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms va

lue of the current in this circuit is 0.680 A, what is the inductance of the inductor?
Physics
2 answers:
Verizon [17]3 years ago
5 0

Answer:

<u>Inductance,L:</u>

"The property of the conductor or the solenoid to generate the electromotive force,emf due to the flow of current,I."

<u>Unit:</u>  henry,H as it is equivalent to, kg.m².sec⁻².A⁻².

Explanation:

<u>Data:</u>

  • Voltage,v=120 v-rms,
  • Frequency,f=1000 Hz,
  • Capacitor, C=2.00 μF,
  • Current,I=0.680 A,

<u>Solution:</u>

We need to calculate the inductance, L of the solenoid inside a circuit,

  1. I=v/z,
  2. Z=√R²+(Lω-1/Cω)²,
  3. putting the values
  4. I=V/√R²+(Lω-1/Cω)²,
  5. 0.680=120/√(100)²+(L×2π×1000-1/2×10⁻⁶×2π×1000)²,
  6. L=35.8×10⁻³H, or<u> L=35.8 mH.⇒</u>Answer
natima [27]3 years ago
4 0

Answer:

The inductance of the inductor is 35.8 mH

Explanation:

Given that,

Voltage = 120-V

Frequency = 1000 Hz

Capacitor C= 2.00\mu F

Current = 0.680 A

We need to calculate the inductance of the inductor

Using formula of current

I = \dfrac{V}{Z}

Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}

Put the value of Z into the formula

I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}

Put the value into the formula

0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}

L=35.8\ mH

Hence, The inductance of the inductor is 35.8 mH

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