Respuesta :
Explanation:
It is given that,
Resistance, [tex]R=9\times 10^2\ \Omega[/tex]
Capacitance, [tex]C=0.25\ mF=0.25\times 10^{-3}\ F[/tex]
Inductance, L = 2.5 H
Frequency, [tex]f=2.4\times 10^2\ Hz[/tex]
Maximum voltage, [tex]V_{max}=1.4\times 10^2\ V[/tex]
(a) Impedance of the circuit is given by :
[tex]Z=\sqrt{R^2+(X_L-X_C)^2}[/tex]...............(1)
[tex]X_L=2\pi fL[/tex]
[tex]X_L=2\pi \times 2.4\times 10^2\times 2.5=3769.91\ \Omega[/tex]
[tex]X_C=\dfrac{1}{2\pi fC}[/tex]
[tex]X_C=\dfrac{1}{2\pi \times 2.4\times 10^2\times 0.25\times 10^{-3}}[/tex]
[tex]X_C=2.65\ \Omega[/tex]
Impedance, [tex]Z=\sqrt{(9\times 10^2)^2+(3769.91-2.65)^2}[/tex]
Z = 3873.27 ohms
(b) Using Ohm's law, [tex]I_{max}=\dfrac{V_{max}}{R}[/tex]
[tex]I_{max}=\dfrac{1.4\times 10^2}{9\times 10^2}[/tex]
[tex]I_{max}=0.15\ A[/tex]
(c) Phase angle is given by :
[tex]\phi=tan^{-1}(\dfrac{X_L-X_C}{R})[/tex]
[tex]\phi=tan^{-1}(\dfrac{3769.91-2.65}{9\times 10^2})[/tex]
[tex]\phi=76.5^{\circ}[/tex]
Hence, this is the required solution.
