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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?

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    \$\begingroup\$ 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. \$\endgroup\$ – MathieuL Jan 20 at 20:40
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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:

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

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