Answer:
L= 2 mH
Step-by-step 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
![E=(1)/(2)LI^2](https://img.qammunity.org/2021/formulas/physics/high-school/uaifshp5gm6gwzkxwq6f6pmtgi6ay0qh22.png)
Where ,L= Inductance
I=Current
E=Energy
Now by putting the values in the above equation
![10^(-5)=(1)/(2)* L* 0.1^2](https://img.qammunity.org/2021/formulas/physics/high-school/guc7c7azep53r3uzkw2fw1fi5g0gsuf17e.png)
![L=(2* 10^(-5))/(0.1^2)\ H](https://img.qammunity.org/2021/formulas/physics/high-school/hjdl2cd8jn5exibasgxgir94rb1etwick4.png)
L=0.002 H
L= 2 mH
We know that frequency f is given as
![2\pi f=(1)/(√(LC))](https://img.qammunity.org/2021/formulas/physics/high-school/q3u5kktoqbhvi17jui502w9re2lldxaa88.png)
C=Capacitance , f=frequency ,L=Inductance
Now by putting the values
![2\pi * 10* 10^3=(1)/(√(0.002* C) )](https://img.qammunity.org/2021/formulas/physics/high-school/wg32gkzkhuaoccihtso6e78ts8mhtj0eo8.png)
![62831.85=(1)/(√(0.002* C))](https://img.qammunity.org/2021/formulas/physics/high-school/gfu5h7kcswukkdbsnoudx0437gl08vowv4.png)
![\sqrt{0.002* C=(1)/(62831.85)](https://img.qammunity.org/2021/formulas/physics/high-school/8o6jzk7ensf4he6w2mc5uozcis2pjsv2uz.png)
![0.002* C=0.0000159^2](https://img.qammunity.org/2021/formulas/physics/high-school/5cmvny3k9eagqudpkj2pzpwdkr6tmrr0xn.png)
![C=(0.0000159^2)/(0.002)\ F](https://img.qammunity.org/2021/formulas/physics/high-school/1tfgufxtxil24ruy9549d0dsenrev66rd6.png)
![C=1.26* 10^(-7)\ F](https://img.qammunity.org/2021/formulas/physics/high-school/kxs0h4xu8glhid8voyfgxfpgr90f0iw96w.png)
Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.