# 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. Commented Jan 20, 2020 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:

$$$$\begin{split} \text{B}&=\Im\left(\frac{1}{\text{R}+\text{j}\omega\text{L}}\right)\\ \\ &=\Im\left(\frac{1}{\text{R}+\text{j}\omega\text{L}}\cdot\frac{\text{R}-\text{j}\omega\text{L}}{\text{R}-\text{j}\omega\text{L}}\right)\\ \\ &=\Im\left(\frac{\text{R}-\text{j}\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}\right)\\ \\ &=\Im\left(\frac{\text{R}}{\text{R}^2+\left(\omega\text{L}\right)^2}-\frac{\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}\cdot\text{j}\right)\\ \\ &=\Im\left(\underbrace{\frac{\text{R}}{\text{R}^2+\left(\omega\text{L}\right)^2}}_{=\space\Re}+\left(\underbrace{-\frac{\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2}}_{=\space\Im}\right)\cdot\text{j}\right)\\ \\ &=-\frac{\omega\text{L}}{\text{R}^2+\left(\omega\text{L}\right)^2} \end{split}\tag3$$$$

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:

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