At what frequency does the inductive reactance of a 42.0 µH inductor equal the capacitive reactance of a 42.0 µF capacitor?
kHz
reactance magnitude = V/i
for inductance
V = L di/dt
if i = a sin wt
di/dt = a w cos wt
and
V = L a w cos wt
magnitude of V/i = Lw = 2 pif L
here L = 42 * 10^-6
so
reactance = 2 pi f (42*10^-6)
for capacitor
C = q/V
V = q/C = (1/C) integral i dt
if i = a sin w t
V = (1/C) (a/w)(-cos wt)
magnitude of reactance =
1/(wC) = 1/(2 pi f C)
= 1/(2 pi f (42*10^-6))
so
1/(2 pi f (42*10^-6)) = 2 pi f (42*10^-6)
1 =( 2 pi f)^2 (42^2) 10^-12
(2 pi f)^2 = 10^12/42^2
2 pi f = 10^6/42 = 2.38*10^6
f = .379 * 10^6
= 379,000 Hz
2 pi f = 10^6/42 = 2.38*10^4
f = .379 * 10^4
= 3790 Hz
The formula to calculate the inductive reactance (XL) of an inductor and the capacitive reactance (XC) of a capacitor is given by:
XL = 2πfL
XC = 1/(2πfC)
where f is the frequency in Hz, L is the inductance in Henries, and C is the capacitance in Farads.
To find the frequency at which the inductive reactance of the inductor is equal to the capacitive reactance of the capacitor, we can set XL equal to XC and solve for f.
2πfL = 1/(2πfC)
Rearranging the equation:
2πfL = 1/(2πfC)
Simplifying:
f = 1 / (2π√(LC))
Substituting the given values of L = 42.0 µH and C = 42.0 µF (convert µH to H and µF to F):
f = 1 / (2π√(42.0 × 10^(-6) H × 42.0 × 10^(-6) F))
Simplifying:
f ≈ 1 / (2π√(1.764 × 10^(-12) H·F))
f ≈ 1 / (2π × 1.328 × 10^(-6) Hz)
Calculating:
f ≈ 238.734 kHz
Therefore, the frequency at which the inductive reactance of the 42.0 µH inductor equals the capacitive reactance of the 42.0 µF capacitor is approximately 238.734 kHz.
To find the frequency at which the inductive reactance of a 42.0 µH inductor equals the capacitive reactance of a 42.0 µF capacitor, we can use the formula for inductive reactance (XL) and capacitive reactance (XC).
The formula for inductive reactance is:
XL = 2πfL
Where XL is the inductive reactance in ohms, f is the frequency in hertz, and L is the inductance in henries.
The formula for capacitive reactance is:
XC = 1 / (2πfC)
Where XC is the capacitive reactance in ohms, f is the frequency in hertz, and C is the capacitance in farads.
To find the frequency at which the inductive reactance equals the capacitive reactance, we can set XL equal to XC and solve for f:
2πfL = 1 / (2πfC)
Now we can substitute the given values:
42.0 µH = 42.0 x 10^-6 H (convert from microhenries to henries)
42.0 µF = 42.0 x 10^-6 F (convert from microfarads to farads)
Substituting these values into the equation:
2πf(42.0 x 10^-6) = 1 / (2πf(42.0 x 10^-6))
Simplifying the equation:
2πf(42.0 x 10^-6)^2 = 1
Now, solve for f:
f = 1 / (2π(42.0 x 10^-6)^2)
Calculating the value:
f ≈ 110.27 kHz
Therefore, the frequency at which the inductive reactance of a 42.0 µH inductor equals the capacitive reactance of a 42.0 µF capacitor is approximately 110.27 kHz.