# How to calculate susceptance of an inductor in series with a resistor

Given the following part of a circuit, I have to find the susceptance.

I know that the susceptance is $$\B=-\frac{1}{reactance}\$$.

I know that the resistor does not have a reactance, however, my textbook's solution is $$\ B=-\frac{wL}{R^2+(wL)^2} \$$. Why does $$\R\$$ come up in the solution?

• I will give you a hint because clearly, you didn't properly analyze this circuit before coming here. Susceptance is the imaginary part of the Admittance. Admittance is the inverse of the impedance. – MathieuL Jan 20 at 20:40

Well, the definition of susceptance is the following:

$$\text{susceptance}=\Im\left(\text{admittance}\right)=\Im\left(\frac{1}{\text{impedance}}\right)\tag1$$

In formula form:

$$\text{B}=\Im\left(\underline{\text{Y}}\right)=\Im\left(\frac{1}{\underline{\text{Z}}}\right)\tag2$$

So, we get:

$$\text{B}=\Im\left(\frac{1}{\text{R}+\text{j}\omega\text{L}}\right)=\Im\left(\frac{\text{R}-\text{j}\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}\right)=$$ $$\Im\left(\frac{\text{R}}{\text{R}^2+\left(\omega\text{L}\right)^2}-\frac{\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}\cdot\text{j}\right)=-\frac{\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}\tag3$$

Because the total impedance of your circuit is a series circuit of a resistor, with impedance $$\\text{R}\$$, and an inductor with impedance $$\\text{j}\omega\text{L}\$$. So:

$$\underline{\text{Z}}=\text{R}+\text{j}\omega\text{L}\tag4$$