Answer:
E) 6.5 A
Explanation:
Given that
L = 40 m H
C= 1.2 m F
Maximum charge on capacitor ,Q= 45 m C
The maximum current I given as
I = Q.ω
ω =angular frequency
[tex]\omega=\dfrac{1}{\sqrt{CL}}[/tex]
By putting the values
[tex]\omega=\dfrac{1}{\sqrt{CL}}[/tex]
[tex]\omega=\dfrac{1}{\sqrt{40\times 10^{-3}\times 1.2 \times 10^{-3}}}[/tex]
ω = 144.33 rad⁻¹
Maximum current
I = 45 x 10⁻³ x 144.33 A
I= 6.49 A
I = 6.5 A
E) 6.5 A