A 22 μF capacitor and a 5.20 mH inductor are connected in series with an open switch. The capacitor is initially charged to 6.70 μC.
What is the angular frequency of the charge oscillations in the capacitor after the switch is closed?
W = _____ rad/s
w=1/LC
CD
To find the angular frequency of the charge oscillations in the capacitor, we can use the formula:
ω = 1 / √(LC)
where ω is the angular frequency, L is the inductance, and C is the capacitance.
Given: capacitor capacitance (C) = 22 μF = 22 x 10^-6 F
inductor inductance (L) = 5.20 mH = 5.20 x 10^-3 H
Plugging these values into the formula:
ω = 1 / √(5.20 x 10^-3 H * 22 x 10^-6 F)
ω = 1 / √(114.4 x 10^-9 H F)
Simplifying the expression under the square root:
ω = 1 / √(114.4 x 10^-15 H^2 F^2)
ω = 1 / (10.7 x 10^-8 H F)
ω = (1 / 10.7) x 10^8 rad/s
Therefore, the angular frequency of the charge oscillations in the capacitor after the switch is closed is:
ω = 9.35 x 10^7 rad/s
To find the angular frequency of the charge oscillations in the capacitor, we need to calculate the value of the equivalent capacitance and the equivalent inductance of the series circuit.
In a series circuit, the total capacitance (Ct) and total inductance (Lt) can be calculated using the following formulas:
1 / Ct = 1 / C1 + 1 / C2 + 1 / C3 + ... (for capacitors in series)
Lt = L1 + L2 + L3 + ... (for inductors in series)
In this case, we have a single capacitor and a single inductor connected in series, so the formulas simplify to:
Ct = C1
Lt = L1
Given that the capacitor has a capacitance of 22 μF (microfarads) and the inductor has an inductance of 5.20 mH (millihenries), we get:
Ct = 22 μF = 22 × 10^-6 F
Lt = 5.20 mH = 5.20 × 10^-3 H
Now we can calculate the angular frequency (ω) using the formula:
ω = 1 / √(Lt * Ct)
Plugging in the values, we get:
ω = 1 / √((5.20 × 10^-3) * (22 × 10^-6))
Calculating this expression gives us the angular frequency (ω) in rad/s.