Question

An inductor is connected to a $15KHz$oscillator. The peak current is $65mA$when the rms voltage is width="36">6.0V. What is the value of the inductance L?

Short answer

The value of inductance$$\begin{gathered} \left(\mathrm{L}\right)=1.37\times {10}^{-3} \\ \approx 1.4\times {10}^{-3} \end{gathered}$$

Step-by-step solution

GIVEN INFORMATION:

Frequency of oscillator $\left(f\right)=15KHz$

peak current$\left({\mathrm{I}}_{\mathrm{o}}\right)=65mA$

RMS voltage $\left(V_{rms}\right)=6.0V$

Here we need to find the inductance of the inductor through the formula $X_L=2\pi fL$where $X_L$is the inductive reactance of the inductor.

EXPLANATION:

We know that ${\mathrm{I}}_{\mathrm{rms}}=\frac{{\mathrm{I}}_{\mathrm{o}}}{\sqrt{2}}$

and $X_L=\frac{V_{rms}}{{\mathrm{I}}_{\mathrm{rms}}}$

$=\frac{6\sqrt{2}}{65\times {10}^{-3}}$ (substituting values)

$=130\Omega$

now, $X_L=2\pi fL$

$L=\frac{X_L}{2\pi f}$

$=\frac{130}{2\pi \times 15000}$ (Substituting values)

$=1.37\times {10}^{-3}H$

$\approx 1.4\times {10}^{-3}H$