91Ó°ÊÓ

We are given that it is an RL circuit.

We have to find the expressions for $I,V_{R}and{V}_L$.

As we know,

$I=\frac{{\mathrm{Î\mu }}_o}{Z}$and $Z=\sqrt{R^2+{X_L}^2}$

then $I=\frac{{\mathrm{Î\mu }}_o}{\sqrt{R^2+{X_L}^2}}$and

we get,$V_R=IR=\frac{{\mathrm{Î\mu }}_{o}R}{\sqrt{R^2+{X_L}^2}}$Also,

$V_L=I{X}_L=\frac{{\mathrm{Î\mu }}_o{X}_L}{\sqrt{R^2+{X_L}^2}}$

Hence, these are the requires equations.

We have to find out $V_R$in limits$\mathrm{Ó\neg }→0and\mathrm{Ó\neg }→∞$.

As,

$V_R=\frac{{\mathrm{Î\mu }}_{o}R}{\sqrt{R^2+{X_L}^2}}$

and if $\mathrm{Ó\neg }→∞$then $V_R→{\mathrm{Î\mu }}_o$.

We are given that output is taken from resistor.

We have to find out if it is low-pass filter or high-pass filter.

As we know that low-pass filter transmits low frequencies and block high frequencies and high-pass filter transmits high frequencies and block low frequencies.

And resistor have low frequency so it will have low-pass filter.

We have to find out the expression for crossover frequency${\mathrm{Ó\neg }}_C$.

As, it is RL circuit

which gives$X_L=2\mathrm{π}×\mathrm{f}×\mathrm{L}$

and also,

$L=\frac{0.15916}{{\mathrm{Ó\neg }}_c}$

On simplifying

we get,${\mathrm{Ó\neg }}_c=\frac{0.15916}{L}$.