Question

5. A series RLC circuit utilizes a 200 ? resistor, a 663 mH inductor, and a 26.5 ?F capacitor. The applied voltage has an amplitude of 50 V and a frequency of 60.0 Hz. Find a) The maximum current and its phase relative to the applied voltage. X_L = ?L = 2?fL ? X_L = 2?(60)(663x10?³) = 250 ? X_C = 1/?C = 1/(2?fC) ? X_C = 1/(2?(60)(26.5x10??)) ? X_C = 100 ? Z = ?R² + (X_L - X_C)² ? Z = ?200² + (250-100)² ? Z = 250 ? I_rms = V_rms/Z ? I_rms = 50/250 ? I_rms = 0.2 A, ? = tan?¹((X_L - X_C)/R) ? ? = 36.86° b) The maximum voltage across the resistor and its phase relative to the current V_max = I_max * R V_max = (0.2)(200) V_max = 40V c) The maximum voltage across the capacitor and its phase relative to the current. V_max = I_max * X_C V_max = (0.2)(100) V_max = 20V d) The maximum voltage across the inductor and its phase relative to the current. V_max = I_max * X_L V_max = (0.2)(250) V_max = 50V

          5. A series RLC circuit utilizes a 200 ? resistor, a 663 mH inductor, and a 26.5 ?F capacitor. The applied voltage has an amplitude of 50 V and a frequency of 60.0 Hz. Find
a) The maximum current and its phase relative to the applied voltage.
X_L = ?L = 2?fL ? X_L = 2?(60)(663x10?³) = 250 ?
X_C = 1/?C = 1/(2?fC) ? X_C = 1/(2?(60)(26.5x10??)) ? X_C = 100 ?
Z = ?R² + (X_L - X_C)² ? Z = ?200² + (250-100)² ? Z = 250 ?
I_rms = V_rms/Z ? I_rms = 50/250 ? I_rms = 0.2 A, ? = tan?¹((X_L - X_C)/R) ? ? = 36.86°
b) The maximum voltage across the resistor and its phase relative to the current
V_max = I_max * R
V_max = (0.2)(200)
V_max = 40V
c) The maximum voltage across the capacitor and its phase relative to the current.
V_max = I_max * X_C
V_max = (0.2)(100)
V_max = 20V
d) The maximum voltage across the inductor and its phase relative to the current.
V_max = I_max * X_L
V_max = (0.2)(250)
V_max = 50V

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5. A series RLC circuit utilizes a 200 ? resistor, a 663 mH inductor, and a 26.5 ?F capacitor. The applied voltage has an amplitude of 50 V and a frequency of 60.0 Hz. Find
a) The maximum current and its phase relative to the applied voltage.
XL = ?L = 2?fL ? XL = 2?(60)(663x10?³) = 250 ?
XC = 1/?C = 1/(2?fC) ? XC = 1/(2?(60)(26.5x10??)) ? XC = 100 ?
Z = ?R² + (XL - XC)² ? Z = ?200² + (250-100)² ? Z = 250 ?
Irms = Vrms/Z ? Irms = 50/250 ? Irms = 0.2 A, ? = tan?¹((XL - XC)/R) ? ? = 36.86°
b) The maximum voltage across the resistor and its phase relative to the current
Vmax = Imax * R
Vmax = (0.2)(200)
Vmax = 40V
c) The maximum voltage across the capacitor and its phase relative to the current.
Vmax = Imax * XC
Vmax = (0.2)(100)
Vmax = 20V
d) The maximum voltage across the inductor and its phase relative to the current.
Vmax = Imax * XL
Vmax = (0.2)(250)
Vmax = 50V

Added by Ruben B.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Sri K

07/31/2023

Step 1

First, we need to find the impedance of the RLC circuit. The impedance is given by the formula: $Z = R + j(X_L - X_C)$ where $R$ is the resistance, $X_L$ is the inductive reactance, and $X_C$ is the capacitive reactance. Show more…

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A series RLC circuit utilizes a 200 Ω resistor, a 663 mH inductor, and a 26.5 μF capacitor. The applied voltage has an amplitude of 50 V and a frequency of 60.0 Hz. Find a) The maximum current and its phase relative to the applied voltage. b) The maximum voltage across the resistor and its phase relative to the current. c) The maximum voltage across the capacitor and its phase relative to the current. d) The maximum voltage across the inductor and its phase relative to the current.

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Transcript

00:01 That's all the given quotient in this quotient based on circuit having the rs circuit which is in the rsc circuit as r l and c are connected in series with the provider with the potential difference so the value of r here is 200 ohm the value of l that is inductance is 2 that is 663 milli henry and the capacitance is 26 .5 micro farad so we can first of all we write that is resistance r is equals to 200 ohm the value of inductance is 663 milli henry or we can write 663 into 10 to the power minus of 3 henry the value of capacitance that is 26 .5 micro farad that is equals to 26 .5 into 10 to the power minus of 6 farad now we have to calculate the current maximum value so for this first of all we calculate the impedance that is the value of z the impedance is given by under root of r square plus xl minus xd whole square here xl and xd are value of xl is equals to 2 pi f into l that is substituting the value we get 2 into 2 pi into f that is here in this quotient the frequency given that is f is equals to 60 hertz and the voltage v is equals to 50 volt now f that is 60 hertz and the value of l that is 663 into 10 to the power minus of 3 on solving this we get the value of x of l is equals to 250 ohm now calculating the value of x of c is equals to 1 upon 2 pi f f into c which is equals to 1 upon 2 pi into f that is 60 into c that is 26 .5 into 10 to the power minus of 6 on solving this we get the value of x of c is equals to 100 ohm now the impedance is equals to under root of r square plus xl minus of xd whole square which is equals to under root of r square that is 200 whole square plus 250 minus of 100 whole square solving this way we get the value of impedance is equals to 250 now the value of current maximum that is i max is equals to v maximum upon impedance that is z on substituting the value we get v max that is 50 divided by 250 that is value of i max which gives 0 .2 ampere that is i maximum that the current maximum value is 0 .2 ampere this…

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