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
25H
We cannot calculate the inductive reactance without knowing the value of the frequency and the formula for inductive reactance (XL = 2πfL).
222,5 micro H
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.
1. Inductance of the coil: Well, it seems like we need to channel our inner algebra skills for this one. Let's start by using the formula for the resonant frequency of an LC circuit:
f = 1 / (2π√(LC))
Since the frequency is given as 500KHz, we can plug in the values:
500KHz = 1 / (2π√(L(455pF)))
Solving for L, we get:
L = (1 / (500KHz * 2π))^2 * 455pF
After doing the math, we find the inductance of the coil.
2. Inductive reactance: Ah, now we're talking about the imaginary parts of complex numbers. The inductive reactance (XL) can be calculated using the formula:
XL = 2πfL
Substituting the given values, we can find the inductive reactance.
Remember, electrical equations can sometimes be shocking, so be mindful of the units and don't get too charged up!
To calculate the inductance of the coil and the inductive reactance, we can use the formulas related to the current and impedance in an AC circuit.
1. Inductance of the coil:
- At resonance, the current has its maximum value. This implies that the reactive components of the circuit (inductive and capacitive reactance) cancel each other out, resulting in a purely resistive circuit.
- At resonance, the impedance of the circuit (Z) is given by Z = R, where R is the resistance of the coil. We can express the impedance of the circuit in terms of inductive reactance (XL) and capacitive reactance (XC) using the following formula:
Z = √(R^2 + (XL - XC)^2)
- Given that the capacitance at resonance is 455pF and the frequency is 500KHz (which can be converted to 500,000Hz), we can calculate the inductive reactance at resonance using the formula:
XC = 1 / (2πfC)
= 1 / (2π * 500,000 * 455 * 10^-12)
- At resonance, XL = XC since they cancel each other out, leading to a purely resistive circuit:
XL = XC
XL = 1 / (2π * 500,000 * 455 * 10^-12)
- Now we can calculate the impedance (Z) at resonance using the formula:
Z = √(R^2 + (XL - XC)^2)
Assume R = 0:
Z = √((XL - XC)^2)
Z = √(XL^2 - 2XLXC + XC^2)
- Since XL = XC at resonance:
Z = √(XL^2 - 2XL^2 + XL^2)
Z = √(XL^2)
Z = XL
- Therefore, we can calculate the inductance of the coil at resonance using the formula for inductive reactance:
XL = 2πfL
L = XL / (2πf)
L = XL / (2π * 500,000)
L = (1 / (2π * 500,000 * 455 * 10^-12)) / (2π * 500,000)
2. Inductive reactance:
- The inductive reactance (XL) in an AC circuit with inductance (L) and frequency (f) is given by:
XL = 2πfL
- Given that the capacitance is reduced to 424pF and the frequency is still 500KHz, we can calculate the new inductive reactance using the formula:
XL = 2πfL
XL = (2π * 500,000 * L)
To summarize:
- To calculate the inductance of the coil, use the formula:
L = (1 / (2π * 500,000 * 455 * 10^-12)) / (2π * 500,000)
- To calculate the new inductive reactance, use the formula:
XL = (2π * 500,000 * L)
Plug in the values and perform the calculations to find the inductance and the new inductive reactance.