Answer:
a)[tex]L=0.00142H[/tex]
b) [tex]C=2.65*10^{-12}[/tex]
Explanation:
From the question we are told that:
Frequency[tex]F=13kHz[/tex]
Current [tex]I=0.14A[/tex]
Capacitor[tex]C_e=1.4*10^{-5}J[/tex]
Generally the equation for Energy in the inductor is mathematically given by
Where L is now subject
[tex]L=\frac{2C_e}{I^2}[/tex]
[tex]L=\frac{2*1.4*10^{-5}}{(0.14)^2}[/tex]
[tex]L=0.00142H[/tex]
Generally the equation for Value of Capacitor is mathematically given by
[tex]C=\frac{1}{(2 \pi f)^2} L[/tex]
[tex]C=\frac{1}{(2 3.142 13*10^3Hz)^2} *0.00142[/tex]
[tex]C=2.65*10^{-12}[/tex]