he impedance of an \(L C R\) circuit is $$ Z=R+\mathrm{j}\left(\omega L-\frac{1}{\omega C}\right) $$ (a) Find \(|Z|\). (b) From the result of part (a) deduce that the impedance has minimum magnitude when $$ \omega=\sqrt{\frac{1}{(L C)}} $$ (c) Deduce that this minimum value is \(R\).

We can find the magnitude of Z using the formula \(|Z|=\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}\). We have: $$ |Z| = \sqrt{R^2 + \left(\omega L - \frac{1}{\omega C}\right)^2} $$

In order to find the condition for the minimum magnitude, we can differentiate |Z| with respect to \(\omega\) and set the derivative equal to zero. This will give us the value of \(\omega\) for which |Z| is minimum. The derivative will be calculated as follows: $$ \frac{d|Z|}{d\omega}=\frac{d}{d\omega}\sqrt{R^2+\left(\omega L-\frac{1}{\omega C}\right)^2} $$ Now, let $$ f(\omega)=R^2+\left(\omega L-\frac{1}{\omega C}\right)^2 $$ Then, $$ \frac{d|Z|}{d\omega}=\frac{d\sqrt{f(\omega)}}{d\omega}=\frac{1}{2\sqrt{f(\omega)}}\frac{df(\omega)}{d\omega} $$ Next, we find \(\frac{df(\omega)}{d\omega}\): $$ \frac{df(\omega)}{d\omega}=2\left(\omega L-\frac{1}{\omega C}\right)\left(L+\frac{1}{\omega^2 C}\right) $$ Set \(\frac{df(\omega)}{d\omega}=0\) and solve for \(\omega\): $$ \omega L - \frac{1}{\omega C} = 0 \implies \omega = \sqrt{\frac{1}{LC}} $$

Now we can plug the value of \(\omega\) we found in Step 2 back into the |Z| equation from Step 1: $$ |Z_{min}| = \sqrt{R^2 + \left(\sqrt{\frac{1}{LC}}L - \frac{1}{\sqrt{\frac{1}{LC}} C}\right)^2} = \sqrt{R^2 + \left(1 - 1\right)^2} = \sqrt{R^2} = R $$ Therefore, the minimum value of |Z| is R.