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?
2 answers:
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>
<u>Solution:</u>
We need to calculate the inductance, L of the solenoid inside a circuit,
- I=v/z,
- Z=√R²+(Lω-1/Cω)²,
- putting the values
- I=V/√R²+(Lω-1/Cω)²,
- 0.680=120/√(100)²+(L×2π×1000-1/2×10⁻⁶×2π×1000)²,
- L=35.8×10⁻³H, or<u> L=35.8 mH.⇒</u>Answer
Answer:
The inductance of the inductor is 35.8 mH
Explanation:
Given that,
Voltage = 120-V
Frequency = 1000 Hz
Capacitor
Current = 0.680 A
We need to calculate the inductance of the inductor
Using formula of current
Put the value of Z into the formula
Put the value into the formula
Hence, The inductance of the inductor is 35.8 mH
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