91Ó°ÊÓ

We have been given that the parallel RLC circuit shown in $FIGURECP32.72$.

We need to show that the current drawn from the emf is

$I={\mathrm{Î\mu }}_o\sqrt{\frac{1}{R^2}+{\left(\frac{1}{\mathrm{Ó\neg }L}-\mathrm{Ó\neg }C\right)}^2}$

We know the formula:

$I^2=I_R^2+{(I_L-I_C)}^2$ (Let this equation be$\left(1\right)$)

where, $I$is the total current in RLC circuit, $I_R$is the current passing through resistor, $I_L$is the current passing through inductor and $I_C$is the current passing through capacitor.

Now, $I=\frac{{\mathrm{Î\mu }}_o}{Z}$, where ${\mathrm{Î\mu }}_o$is emf and $Z$is the impedance.

$I_R=\frac{{\mathrm{Î\mu }}_o}{R}$, where $R$is resistance.

$I_L=\frac{{\mathrm{Î\mu }}_o}{\mathrm{Ó\neg }L}$, where$L$is inductance.

$I_C=\frac{{\mathrm{Î\mu }}_o}{{\displaystyle \frac{1}{\mathrm{Ó\neg }C}}}$, where $C$is capacitance.

Now, substitute the values of $I,I_R,I_L,I_C$in equation $\left(1\right)$, we get:

$\left(\frac{{\mathrm{Î\mu }}_o}{Z}\right)={\left(\frac{{\mathrm{Î\mu }}_o}{R}\right)}^2+{\left(\frac{{\mathrm{Î\mu }}_o}{\mathrm{Ó\neg }L}-\frac{{\mathrm{Î\mu }}_o}{{\displaystyle \frac{1}{\mathrm{Ó\neg }C}}}\right)}^2$

Rearranging the terms, we get:

$Z={\left[\frac{1}{R^2}+{\left(\frac{1}{\mathrm{Ó\neg }L}-\mathrm{Ó\neg }C\right)}^2\right]}^{\raisebox{1ex}{$-1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Now substitute $Z=\frac{{\mathrm{Î\mu }}_o}{I}$, we get:

$I={\mathrm{Î\mu }}_o\sqrt{\frac{1}{R^2}+{\left(\frac{1}{\mathrm{Ó\neg }L}-\mathrm{Ó\neg }C\right)}^2}$

Hence Proved.$I_R=\frac{{\mathrm{Î\mu }}_o}{R}$

We have been given that the parallel RLC circuit shown in $FIGURECP32.72$.

We need to find$I$in the limits$\mathrm{Ó\neg }→0$and$\mathrm{Ó\neg }→∞$.

We know that when $\mathrm{Ó\neg }→0\raisebox{1ex}{$rad$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$, then $\mathrm{Ó\neg }L→0\mathrm{Î\copyright }$

If $\mathrm{Ó\neg }L→0\mathrm{Î\copyright }$, this means there is no hinderance in the path of current.

So, we can say$\mathrm{ι}→∞$.

We have been given that the parallel RLC circuit shown in $FIGURECP32.72$.

We need to find the frequency for which$I$is minimum.

The frequency for $I$is minimum when:

$\frac{dI}{d\mathrm{Ó\neg }}=0$, where $I={\mathrm{Î\mu }}_o\sqrt{\frac{1}{R^2}+{\left(\frac{1}{\mathrm{Ó\neg }L}-\mathrm{Ó\neg }C\right)}^2}$ (from Part(a))

$\frac{dI}{d\mathrm{Ó\neg }}=\frac{{\mathrm{Î\mu }}_o\left({\displaystyle \frac{1}{2}}\right)2\left({\displaystyle \frac{1}{\mathrm{Ó\neg }L}}-\mathrm{Ó\neg }C\right)\left(-{\displaystyle \frac{1}{{\mathrm{Ó\neg }}^{2}L}}-C\right)}{2\sqrt{{\displaystyle \frac{1}{R^2}}+{\left({\displaystyle \frac{1}{\mathrm{Ó\neg }L}}-\mathrm{Ó\neg }C\right)}^2}}$

Now substitute $\frac{dI}{d\mathrm{Ó\neg }}=0$, we get:

$0=\left(\frac{1}{\mathrm{Ó\neg }L}-\mathrm{Ó\neg }C\right)\left(-\frac{1}{{\mathrm{Ó\neg }}^{2}L}-C\right)$

$0=\frac{1}{\mathrm{Ó\neg }L}-\mathrm{Ó\neg }C$

role="math" localid="1650358705411" $\mathrm{Ó\neg }C=\frac{1}{\mathrm{Ó\neg }L}$

${\mathrm{Ó\neg }}^2=\frac{1}{LC}$

${\mathrm{Ó\neg }}_o=\frac{1}{\sqrt{LC}}$

The frequency for which$I$is minimum is${\mathrm{Ó\neg }}_o=\frac{1}{\sqrt{LC}}$.

We have been given that the parallel RLC circuit shown in $FIGURECP32.72$.

We need to sketch a graph of$I$versus$\mathrm{Ó\neg }$.

Graph of $I$versus $\mathrm{Ó\neg }$