An inductor (L = 405 mH), a capacitor (C = 4.43 µF), and a resistor (R = 400 ) are connected in series. A 50.0 Hz AC generator produces a peak current of 250 mA in the circuit.
(a) Calculate the required peak voltage ΔVmax.
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(b) Determine the phase angle by which the current the applied voltage.
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See previous post: Fri,11-29-13,8:15 PM.
To calculate the required peak voltage (ΔVmax) in the circuit, we can use the formulas for reactive components (inductor and capacitor) in an AC circuit.
(a) To find ΔVmax, we need to find the impedance of the circuit, which is the total opposition to the flow of current. The impedance (Z) in an AC circuit is given by:
Z = √((R^2 + (XL - XC)^2))
Where:
R = resistance of the circuit (400 Ω)
XL = inductive reactance (ωL), where ω is the angular frequency and L is the inductance (405 mH = 0.405 H)
XC = capacitive reactance (1 / ωC), where C is the capacitance (4.43 µF = 4.43 x 10^(-6) F)
The angular frequency (ω) is given by:
ω = 2πf
Where:
f = frequency of the AC generator (50 Hz)
Substituting the values into the formulas, we get:
ω = 2π(50 Hz) = 100π rad/s
XL = (100π rad/s)(0.405 H) = 40.5π Ω
XC = 1 / (100π rad/s)(4.43 x 10^(-6) F) = 1 / (443π x 10^(-3)) Ω
Now, we can calculate the impedance (Z):
Z = √((400 Ω)^2 + (40.5π Ω - (1 / 443π x 10^(-3)) Ω)^2)
Finally, to calculate ΔVmax, we use Ohm's law:
ΔVmax = Ipeak * Z
Where:
Ipeak = peak current in the circuit (250 mA = 0.250 A)
Substituting the values, we can calculate ΔVmax.
(b) To determine the phase angle by which the current lags or leads the applied voltage, we need to find the phase angle (θ) between the voltage and current in the circuit.
θ = arctan((XL - XC) / R)
Substituting the values, we can calculate the phase angle (θ).
Please note that all the calculations are based on the given values and formulas for impedance, Ohm's law, and the phase angle in an AC circuit.