Answer:
Explanation:
we have given E(t)=120 sin(12t)
R=5 ohm
L=0.2 H
ω=12 ( from expression of E)
[tex]X_L=0.2\times 12=2.4[/tex] ohm
[tex]X_C=\frac{1}{\omega \times C}=\frac{1}{12\times 0.043}=1.9379\ ohm[/tex]
[tex]Z=\sqrt{R^2+\left ( \omega L-\frac{1}{\omega C} \right )^2}[/tex]
[tex]Z=\sqrt{5^2+\left ( \2.4-1.9379 )^2}[/tex]
=5.021 ohm
so amplitude of current = [tex]\frac{v}{z}=\frac{120}{5.021}=23.89[/tex]
