Answer:
L= 2 mH
[tex]C=1.26\times 10^{-7}\ F[/tex]
Explanation:
Given that
Frequency , f= 10 kHz
Maximum current ,I = 0.1 A
Maximum energy stored ,E= 1 x 10⁻⁵ J
The maximum energy stored in the inductor is given as follows
[tex]E=\dfrac{1}{2}LI^2[/tex]
Where ,L= Inductance
I=Current
E=Energy
Now by putting the values in the above equation
[tex]10^{-5}=\dfrac{1}{2}\times L\times 0.1^2[/tex]
[tex]L=\dfrac{2\times 10^{-5}}{0.1^2}\ H[/tex]
L=0.002 H
L= 2 mH
We know that frequency f is given as
[tex]2\pi f=\dfrac{1}{\sqrt{LC}}[/tex]
C=Capacitance , f=frequency ,L=Inductance
Now by putting the values
[tex]2\pi \times 10\times 10^3=\dfrac{1}{\sqrt{0.002\times C} }[/tex]
[tex]62831.85=\dfrac{1}{\sqrt{0.002\times C}}[/tex]
[tex]\sqrt{0.002\times C=\dfrac{1}{62831.85}[/tex]
[tex]0.002\times C=0.0000159^2[/tex]
[tex]C=\dfrac{0.0000159^2}{0.002}\ F[/tex]
[tex]C=1.26\times 10^{-7}\ F[/tex]
Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.
Given the following data:
Frequency = 10 kHz
Maximum current = 0.10 Amps
Maximum energy = [tex]1.0 \times 10^{-5}[/tex] Joules
To find the values of inductance and capacitance that must be used for this LC circuit:
For the inductance:
Mathematically, the maximum energy stored in an inductor is given by the formula:
[tex]E = \frac{1}{2} LI^2[/tex]
