Respuesta :
Answer:
Explanation:
[tex]a)\: X_C=\frac{1}{C\omega} =\frac{1}{0.073*65} =0.21 \:\Omega\\\\b)\: X_L=L\omega=1.6*65=104\:\Omega\\\\c)Z=\sqrt{R^2+(X_L-X_C)^2}=\sqrt{35^2+(104-0.21)^2}=109.53\:\Omega\\\\d)\: I_0=\frac{V_0}{Z} =\frac{110}{109.53} =1.004 \:A\\\\e)\;I_{rms}=\frac{I_0}{\sqrt2} =\frac{1.004}{\sqrt2} =0.71\:A\\\\f)\:V_{rms}=\frac{V_0}{\sqrt2} =\frac{110}{\sqrt2} =77.78\:V\\\\\cos(\phi)=\frac{R}{Z} =\frac{35}{109.53} =0.32\\\\\phi =\arccos (0.319547)=71.36^\circ[/tex]
[tex]P_{av}=V_{rms}I_{rms}\cos(\phi)=77.78*0.71*0.32=17.67\:W[/tex]
Answer:
(a) 0.21 Ω
(b) 104 Ω
(c) 109.5 Ω
(d) 1.0 A
(e) 0.7072 A
(f) 17.66 W
Explanation:
Given, an AC circuit containing a resistor, a capacitor and an inductor and;
R = Resistance of the resistor = 35 Ω
L = Inductance of the inductor = 1.6H
C = Capacitance of the capacitor = 0.073F
V = V₀ sin (ωt) -------------------(i)
Where;
V₀ = Amplitude voltage = 110V
ω = angular frequency = 65rad/s [Note: ω = 2 [tex]\pi[/tex] f, where f = frequency]
Now;
(a)The capacitive reactance ([tex]X_{C}[/tex]) is the opposition given to the flow of current due to an amount of capacitance (C) and frequency (f) of current in a capacitor. It is measured in ohms (Ω) and it is given by;
[tex]X_{C}[/tex] = [tex]\frac{1}{2\pi f C}[/tex] = [tex]\frac{1}{w C}[/tex] ------------------- (ii)
Substitute the appropriate values of the variables in equation (ii)
[tex]X_{C}[/tex] = [tex]\frac{1}{65 * 0.073}[/tex]
[tex]X_{C}[/tex] = [tex]\frac{1}{4.745}[/tex]
[tex]X_{C}[/tex] = 0.21Ω
Therefore, the capacitive reactance of the circuit in ohms is 0.21
(b)The inductive reactance ([tex]X_{L}[/tex]) is the opposition given to the flow of current due to an amount of inductance (L) and frequency (f) of current in an inductor. It is measured in ohms (Ω) and it is given by;
[tex]X_{L}[/tex] = 2 [tex]\pi[/tex] f L = ω L ------------------- (iii)
Substitute for the appropriate values of the variables in equation (iii)
[tex]X_{L}[/tex] = 65 x 1.6
[tex]X_{L}[/tex] = 104Ω
Therefore, the inductive reactance of the circuit in ohms is 104
(c)The impedance (Z) is the resultant or effective resistance in an AC due to the combined effects of the reactances and resistance in the circuit. In an RLC circuit, the impedance is given by;
Z = [tex]\sqrt{[R^{2} + (X_{L} - X_{C})^2]}[/tex] -----------------(iv)
Substitute for the appropriate values of the variables in equation (iv)
Z = [tex]\sqrt{[35^{2} + (104 - 0.21)^2]}[/tex]
Z = [tex]\sqrt{[35^{2} + (103.8)^2]}[/tex]
Z = [tex]\sqrt{[1225 + 10774.44]}[/tex]
Z = [tex]\sqrt{11999.44}[/tex]
Z = 109.5
Therefore, the impedance of the circuit in ohms is 109.5
(d) The maximum current ([tex]I_{MAX}[/tex]) in the circuit is given as follows;
[tex]I_{MAX}[/tex] = [tex]\frac{V_{MAX}}{Z}[/tex] --------------------(i)
Where;
[tex]V_{MAX}[/tex] = V₀ = Maximum voltage or Amplitude voltage = 110V
Z = impedance of the circuit = 109.5Ω
Substitute these values into equation (v) as follows;
[tex]I_{MAX}[/tex] = [tex]\frac{110}{109.5}[/tex]
[tex]I_{MAX}[/tex] = 1.0A
Therefore, the maximum current in the circuit, in amperes, is 1.0
(e) The root mean square(rms) current, [tex]I_{rms}[/tex], in a circuit, is the effective value of the current in the circuit and is given by;
[tex]I_{rms}[/tex] = [tex]\frac{I_{0} }{\sqrt{2} }[/tex] ---------------(vi)
Where;
[tex]I_{0}[/tex] = peak or maximum current = [tex]I_{MAX}[/tex] = 1.0A [as calculated above]
Substitute this value into equation (vi) as follows;
[tex]I_{rms}[/tex] = [tex]\frac{1.0 }{\sqrt{2} }[/tex]
[tex]I_{rms}[/tex] = [tex]\frac{1.0 }{1.414}[/tex]
[tex]I_{rms}[/tex] = 0.7072A
Therefore, the rms current in the circuit, in amperes is 0.7072
(f) The average power ([tex]P_{avg}[/tex]) delivered to the circuit is given by;
[tex]P_{avg}[/tex] = [tex]\frac{V_{rms} ^{2} * R}{Z^2}[/tex] -------------------(vii)
Where;
[tex]V_{rms}[/tex] = rms voltage = [tex]\frac{V_{0} }{\sqrt{2} }[/tex] = [tex]\frac{110}{\sqrt{2} }[/tex] = [tex]\frac{110 }{1.414}[/tex] = 77.79V
Now substitute the values of [tex]V_{rms}[/tex], R and Z into equation (vii) as follows;
[tex]P_{avg}[/tex] = [tex]\frac{77.79 ^{2} * 35}{109.5^2}[/tex]
Solve for [tex]P_{avg}[/tex];
[tex]P_{avg}[/tex] = 17.66W
Therefore, the average power delivered to the circuit in watts is 17.66W
