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.
V
(b) Determine the phase angle by which the current the applied voltage.
°
a. Xl = 2pi*F*l = 6.28*50*0.405=127.2 Ohms
Xc = 1/(6.28*50*4.43*10-6) = -718.9 Ohms
tan A = (Xl-Xc)/R
tan A = (127.2-718.9)/400 = -1.47925
A = -55.94o
Z=R/cosA = 400/cos(-55.94) = 714.2 Ohms[-55.94o]= The Impedance.
Vmax = Imax * Z=0.25 * 714.2 =178.6 Volts.
b. The negative impedance angle means that the circuit is capacitive. Therefore, the current leads the applied
voltage by 55.94. Imax = 0.25A[55.94o].
To calculate the required peak voltage (ΔVmax) in the given circuit, we need to use the formula:
ΔVmax = I * Z
where
ΔVmax is the peak voltage,
I is the peak current, and
Z is the impedance.
To find the impedance (Z) of the series circuit, we need to use the formula:
Z = √(R^2 + (XL - XC)^2)
where
R is the resistance,
XL is the inductive reactance, and
XC is the capacitive reactance.
The formula for inductive reactance (XL) is:
XL = 2πfL
where
f is the frequency and
L is the inductance.
The formula for capacitive reactance (XC) is:
XC = 1/(2πfC)
Let's calculate the required peak voltage ΔVmax:
Given:
L = 405 mH = 0.405 H
C = 4.43 µF = 4.43 × 10^(-6) F
R = 400 Ω
I = 250 mA = 0.250 A
f = 50.0 Hz
Step 1: Calculate the inductive reactance (XL).
XL = 2πfL
XL = 2π * 50.0 Hz * 0.405 H
XL = 127.23 Ω
Step 2: Calculate the capacitive reactance (XC).
XC = 1 / (2πfC)
XC = 1 / (2π * 50.0 Hz * 4.43 × 10^(-6) F)
XC = 717.27 Ω
Step 3: Calculate the impedance (Z).
Z = √(R^2 + (XL - XC)^2)
Z = √(400 Ω^2 + (127.23 Ω - 717.27 Ω)^2)
Z = √(400 Ω^2 + (-590.04 Ω)^2)
Z = √(400 Ω^2 + 348119.60 Ω^2)
Z = √(348519.60 Ω^2)
Z = 590.20 Ω
Step 4: Calculate the required peak voltage (ΔVmax) using the formula.
ΔVmax = I * Z
ΔVmax = 0.250 A * 590.20 Ω
ΔVmax = 147.55 V
So, the required peak voltage ΔVmax is 147.55 V.
To determine the phase angle by which the current lags or leads the applied voltage, we need to use the formula:
θ = arctan((XL - XC) / R)
where θ is the phase angle.
Let's calculate the phase angle θ:
θ = arctan((XL - XC) / R)
θ = arctan((-590.04 Ω) / 400 Ω)
θ = arctan(-1.48)
Using a calculator, we find:
θ ≈ -55.89° (rounded to two decimal places)
So, the phase angle by which the current lags the applied voltage is approximately -55.89°.