Respuesta :
Explanation:
It is given that,
Resistance, R = 203 ohms
Inductance, [tex]L=0.396\ H[/tex]
Capacitance, [tex]C=4.99\ \mu F=4.99\times 10^{-6}\ F[/tex]
Voltage, V = 3 V
Angular frequency, [tex]\omega=400\ rad/s[/tex]
Impedance is given by :
[tex]Z=\sqrt{R^2(X_L-X_C})^2[/tex]
[tex]Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega c}})^2[/tex]
[tex]Z=\sqrt{203^2+ (400\times 0.396-\dfrac{1}{400\times 4.99\times 10^{-6}}})^2[/tex]
R = Z = 398.22 ohms
Current can be calculated using Ohm's law as :
[tex]I=\dfrac{V}{R}[/tex]
[tex]I=\dfrac{V}{Z}[/tex]
[tex]I=\dfrac{3}{398.22}[/tex]
I = 0.00753
or
[tex]I=7.53\times 10^{-3}\ A[/tex]
So, the current amplitude of the circuit is [tex]7.53\times 10^{-3}\ A[/tex]. Hence, this is the required solution.
Answer:
[tex]7.53\times 10^{-3}[/tex] A
Explanation:
w = angular frequency = 400 rad/s
[tex]V_{o}[/tex] = Amplitude of the source = 3.00 Volts
L = Inductance of Inductor = 0.396 H
[tex]X_{L}[/tex] = Inductive reactance = wL = 400 x 0.396 = 158.4 ohm
C = capacitance of the capacitor = 4.99 x 10⁻⁶ F
[tex]X_{C}[/tex] = Capacitive reactance = [tex]\frac{1}{wC}[/tex] = [tex]\frac{1}{(400)(4.99\times 10^{-6})}[/tex] = 501 ohm
R = resistance of the resistor = 203 ohm
Impedance of the circuit is given as
[tex]z = \sqrt{R^{2}+(X_{L} - X_{C})^{2}}[/tex]
[tex]z = \sqrt{203^{2}+(158.4 - 501)^{2}}[/tex]
[tex]z = 398.23 [/tex] ohm
[tex]i_{o}[/tex] = Amplitude of the current
Amplitude of the current is given as
[tex]i_{o} = \frac{V_{o}}{z}[/tex]
[tex]i_{o} = \frac{3}{398.23}[/tex]
[tex]i_{o} =7.53\times 10^{-3}[/tex] A
Rms Current flowing is given as
A
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