Question

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 value of the current in this circuit is 0.680 A, what is the inductance of the inductor?

Answer 1

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

Answer 2

Answer:

Inductance,L:

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

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

Explanation:

Data:

• Voltage,v=120 v-rms,

• Frequency,f=1000 Hz,

• Capacitor, C=2.00 μF,

• Current,I=0.680 A,

Solution:

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 L=35.8 mH.⇒Answer