Answer:
(G) 75.11 ohm
(H) 0.08 A
(I) 46.2 degree
Explanation:
R = 52 ohm
L = 4.8 m H = 4.8 x 106-3 H
C = 330 nF = 330 x 10^-9 F
Vo = 6 V
(G)
f = 5000 Hz
Let the impedance is Z.
[tex]X_{L}= 2 \pi fL = 2 \times 3.14\times 5000\times 4.8\times 10^{-3}=150.72 ohm[/tex]
[tex]X_{c}= \frac{1}{2 \pi fC}=\frac{1}{2\times 3.14\times 5000\times 330\times 10^{-9}}=96.51 ohm[/tex]
[tex]Z=\sqrt{R^{2}+\left ( X_{L}-X_{c} \right )^{2}}[/tex]
[tex]Z=\sqrt{52^{2}+\left (150.72-96.51)^{2}}=75.11 ohm[/tex]
(H) Let Io be the peak current
[tex]I_{0}=\frac{V_{0}}{Z}=\frac{6}{75.11}=0.0798 A = 0.08 A[/tex]
(I) Let Ф be the phase angle
[tex]tan\phi = \frac{X_{L}-X_{C}}{R}[/tex]
[tex]tan\phi =\frac{150.72-96.51}}{52}=1.0425[/tex]
Ф = 46.2 degree
