1. A 26.0-Ω resistor, a 12.0-µF capacitor, and a 17.0-mH inductor are connected in series with a 150-V generator.
(a) At what frequency is the current a maximum?
2.A series circuit contains only a resistor and an inductor. The voltage V of the generator is fixed. If R = 20 Ω and L = 2.8 mH, find the frequency at which the current is one-fifth its value at zero frequency?
1) Maximum current occurs at resonant frequency
f = sqrt(LC)/(2 pi)
2) Compute the angular frequency w for which the impedance magnitude is increased by a factor of 5.
I = (1/5)(V/R) = V/|Z|
= V/sqrt[R^2 + (wL^2])
V cancels out; solve for w.
Then use
f = 2 pi w
Can you plug in the numbers for #1 please? I'm still a little confused.
Sorry about that. The LC term should have been in the denominator, for problem #1.
To answer these questions, we need to use the concepts of impedance and reactance in electrical circuits.
1. To find the frequency at which the current is a maximum in a series circuit with a resistor, capacitor, and inductor, we need to determine the impedance at different frequencies and find where it is minimized.
The impedance (Z) in a series circuit is given by the formula:
Z = √(R^2 + (Xl - Xc)^2)
where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance.
At the frequency where the current is a maximum, the impedance is at a minimum. Since we are given the values of the resistor (R = 26.0 Ω), capacitor (C = 12.0 µF), and inductor (L = 17.0 mH), we can calculate both Xl and Xc using the formulas:
Xl = 2πfL and Xc = 1 / (2πfC)
where f is the frequency.
Now, we can solve the equation for impedance using the given values and find the frequency at which it is minimized. To do this, we can iterate through different frequency values and calculate the impedance using the formula mentioned above. The frequency with the minimum impedance corresponds to the frequency at which the current is a maximum.
2. In a series circuit containing only a resistor and an inductor, the impedance is given by Z = √(R^2 + Xl^2). The reactance of the inductor is Xl = 2πfL, where f is the frequency, and L is the inductance.
We are given the values of the resistor (R = 20 Ω) and inductor (L = 2.8 mH). To find the frequency at which the current is one-fifth its value at zero frequency, we need to calculate the impedance at zero frequency (Z0) and at the desired frequency (Zf). Then we can solve the equation Zf = Z0/5 for the frequency.
To find Z0, we use the formula Z0 = √(R^2 + Xl0^2), where Xl0 is the reactance of the inductor at zero frequency.
Substituting Xl0 = 0 into the equation, we get Z0 = R.
Now, we can solve Zf = Z0/5 for the frequency by substituting Z0 = R and Zf = √(R^2 + Xlf^2) into the equation. Rearranging the equation, we find:
(R^2 + Xlf^2) = (R/5)^2
Xlf^2 = (R/5)^2 - R^2
(fL)^2 = (R/5)^2 - R^2
f^2 = ((R/5)^2 - R^2)/L^2
f = √(((R/5)^2 - R^2)/L^2)