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

The inductance of the inductor is 35.8 mH

Step-by-step 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 = (V)/(Z)`

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

Put the value of Z into the formula

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

Put the value into the formula

`0.680=\frac{120}{\sqrt{(100)^2+(L*2\pi*1000-(1)/(2*10^(-6)*2\pi*1000))^2}}`

`L=35.8\ mH`

Hence, The inductance of the inductor is 35.8 mH

Answer 2

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.

Step-by-step 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