Use a phasor diagram to analyze the RL circuit of FIGURE P$32.49$. In particular,

a. Find expressions for $I,V_{R}and{V}_L$.

b. What is VR in the limits $\omega \to 0$and $\omega \to \infty$?

c. If the output is taken from the resistor, is this a low-pass or a high-pass filter? Explain.

d. Find an expression for the crossover frequency ${\omega }_C$.

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,

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

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

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

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

Hence, these are the requires equations.

We have to find out $V_R$in limits$\omega \to 0and\omega \to \infty$.

As,

$V_R=\frac{{\epsilon }_{o}R}{\sqrt{R^2+{X_L}^2}}$

and if $\omega \to \infty$then $V_R\to {\epsilon }_o$.

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${\omega }_C$.

As, it is RL circuit

which gives$X_L=2\mathrm{\pi }\times \mathrm{f}\times \mathrm{L}$

and also,

$L=\frac{0.15916}{{\omega }_c}$

On simplifying

we get,${\omega }_c=\frac{0.15916}{L}$.