A series circuit contains only a resistor and an inductor. The voltage V of the generator is fixed. If R = 15 and L = 5.6 mH, find the frequency at which the current is one-half its value at zero frequency. ?? Hz
Thanks!
the cutoff frequency is r/(2pi L) = 15/(2pi*5.6 * 10^-3) = 426 Hz
To find the frequency at which the current is one-half its value at zero frequency in a series circuit with a resistor and an inductor, we can use the formula for the impedance of an inductor in an AC circuit.
The impedance of an inductor, denoted as ZL, is given by the formula:
ZL = jωL
Here, ω represents the angular frequency in radians per second, and L is the inductance in henries.
In a series circuit, the total impedance, Z, is the sum of the impedance of the resistor, ZR, and the impedance of the inductor, ZL. Mathematically, it can be expressed as:
Z = ZR + ZL
The impedance of the resistor, ZR, is equal to the resistance R.
So, substituting the values given in the question, we have:
ZR = R = 15 Ω
L = 5.6 mH = 5.6 × 10^(-3) H
Now, we need to find the frequency at which the current is one-half its value at zero frequency. Let's call this frequency f.
At zero frequency (DC), the impedance of the inductor becomes zero since the angular frequency ω is equal to zero. Therefore, the total impedance at zero frequency is equal to the resistance R.
So, Z = ZR = R = 15 Ω
Now, at the frequency f where the current is one-half its value at zero frequency, the absolute value of the total impedance Z is equal to half of ZR. Mathematically, it can be written as:
|Z| = |ZR|/2
Since the impedance of the inductor is given by ZL = jωL, we can write:
|ZL| = ωL
Substituting the values, we have:
|Z| = |ZR + ZL| = 15 Ω/2
|ZL| = ωL
To solve for f, we need to equate these two expressions and solve for ω.
15 Ω/2 = ω × (5.6 × 10^(-3) H)
Simplifying, we find:
ω = (15 Ω/2) / (5.6 × 10^(-3) H) = 5357 rad/s
Now, to find the frequency f in hertz, we need to convert ω from rad/s to Hz. The conversion formula is:
f = ω / (2π)
Substituting the value of ω, we get:
f = 5357 rad/s / (2π)
Evaluating this expression using π ≈ 3.14159, we find:
f ≈ 853.78 Hz
Therefore, the frequency at which the current is one-half its value at zero frequency is approximately 853.78 Hz.