Question

A voltage at a frequency of 500KHz is maintained across a coil with a variable capacitor .when the capacitor is set to 455pF, the current has its maximum value, but when the capacitance is reduced to 424pF , the current is 0,707 of its maximum value.
Calculate: 1. Inductance of the coil
2.inductive reactive

Answer 1

25H

Answer 2

222,5 micro H

Answer 3

We cannot calculate the inductive reactance without knowing the value of the frequency and the formula for inductive reactance (XL = 2πfL).

Answer 4

Using the given information, we can solve for the inductance of the coil and the inductive reactance as follows:

1. To find the inductance of the coil, we can use the resonant frequency formula:
fr = 1 / (2π√(LC))
where fr is the resonant frequency, L is the inductance, and C is the capacitance.

At resonance, the frequency is 500 kHz and the capacitance is 455 pF, so we have:
500 kHz = 1 / (2π√(L(455×10^-12 F)))
Squaring both sides and rearranging, we get:
L = (1 / ((2π × 500 kHz)^2 × 455×10^-12 F))
L ≈ 25.182 μH

Therefore, the inductance of the coil is approximately 25.182 μH.

2. To find the inductive reactance at 424 pF, we use the formula:
XL = 2πfL
where f is the frequency and L is the inductance.

At 500 kHz and 25.182 μH, we have:
XL1 = 2π × 500 kHz × 25.182 μH
XL1 ≈ 79.294 Ω

At 500 kHz and 424 pF, we can calculate the new resonant frequency as follows:
fr' = 1 / (2π√(L(424×10^-12 F)))
fr' ≈ 680.419 kHz

Using this new frequency, we can find the inductive reactance at 424 pF:
XL2 = 2π × 680.419 kHz × 25.182 μH
XL2 ≈ 284.212 Ω

Finally, we can use the given information that the current at 424 pF is 0.707 of its maximum value to find the maximum current:
0.707I_max = I(424 pF)
I_max = I(455 pF)

Let's assume the maximum current is Imax, then we have:
0.707Imax = Imax × (XL2 / XL1)
0.707 = XL2 / XL1
XL2 = 0.707 × XL1

Substituting the calculated values, we get:
284.212 Ω = 0.707 × 79.294 Ω

Therefore, the inductive reactance at 424 pF is approximately 284.212 Ω.