Question

73. II For your final exam in electronics, you're asked to build an LC circuit that oscillates at $10\text{kHz}$. In addition, the maximum current must be $0.10\text{A}$and the maximum energy stored in the capacitor must be $1.0\times {10}^{-5}\text{J}$. What values of inductance and capacitance must you use?

Short answer

Part (a) The inductance of the inductor is $2.0\text{mH}$

Part (b) The capacitance of the capacitor is$0.13\mu \text{F}$

Step-by-step solution

Step 1. 

The energy stored in the inductor.

$E=\frac{1}{2}L{I}^2$

Here, L is the inductance of the inductor and I is the maximum current in the inductor.

$\begin{array}{l}E=\frac{1}{2}L{I}^2\\ 1.0\times {10}^{-5}\text{J}=\frac{1}{2}\left(L\right){\left(0.10\text{A}\right)}^2\\ L=0.002\text{H}\left(\frac{1000\text{mH}}{1\text{H}}\right)\\ L=2.0\text{mH}\end{array}$

Hence, the inductance of the inductor is$2.0\text{mH}$

Step 2.

The expression of the frequency with which the RC-oscillator oscillates:

$f=\frac{1}{2\pi \sqrt{LC}}$

Here, is the capacitance of the capacitor and is the inductance of the inductor. Rearrange the above equation.

$\begin{array}{l}C=\frac{1}{4{\pi }^2{\left(f\right)}^{2}L}\\ C=\frac{1}{4{\pi }^2{\left(10\text{kHz}\left(\frac{1000\text{Hz}}{1\text{kHz}}\right)\right)}^2\left(2.0\times {10}^{-3}\text{H}\right)}\\ =\left(0.13\times {10}^{-6}\text{F}\right)\left(\frac{1\mu \text{F}}{{10}^{-6}\text{F}}\right)\\ =0.13\mu \text{F}\end{array}$

Therefore, the capacitance of the capacitor is$0.13\mu \text{F}$