Question

72. An electric oscillator is made with a $0.10\mu \text{F}$capacitor and a $1.0\text{mH}$inductor. The capacitor is initially charged to$5.0\text{V}$ . What is the maximum current through the inductor as the circuit oscillates?

Short answer

The current passing through the circuit is 0.05 A.

Step-by-step solution

Step 1. Given information

The value of inductance, and capacitance

$\begin{array}{l}L=\left(1.0\text{mH}\right)\left(\frac{1\text{H}}{1000\text{mH}}\right)=0.001\text{H}\\ C=\left(0.10\mu \text{F}\right)\left(\frac{1\times {10}^{-6}\text{F}}{1\mu \text{F}}\right)=1.0\times {10}^7\text{F}\end{array}$

Step 2. Explanation

Equate the potential energy stored in the capacitor is equal to potential energy stored in inductor.

$\begin{array}{l}{U}_C=U_L\\ \frac{1}{2}L{I}^2=\frac{1}{2}C{\left(\Delta V\right)}^2\\ I=\Delta V\sqrt{\frac{C}{L}}\\ I=(25\text{V})\sqrt{\frac{\sqrt{\left(8.0\text{pF}\right)\left(\frac{1\times {10}^{-12}\text{F}}{1\text{pF}}\right)}}{\left(20\text{mH}\right)\left(\frac{1\text{H}}{1000\text{mH}}\right)}}\\ I=5\times {10}^{-4}\text{A}\end{array}$

Therefore, the current passing through the circuit is $0.05\text{A}$