Question

An inductor, a capacitor, and a resistor are all connected in series across an ac source. If the resistance, inductance, and capacitance are all doubled, by what factor does each of the following quantities change? Indicate whether they increase or decrease: (a) the resonance angular frequency; (b) the inductive reactance; (c) the capacitive reactance. (d) Does the impedance double?

Short answer

a) Resonance angular frequency is halved

b) The inductive reactance doubles

c) Capacitive reactance halves and

d) Z increases without doubling.

Step-by-step solution

Step-1: Formulas used  

Resonance angular frequency is given by $\mathit{\omega }\mathbf{=}\frac{\mathbf{1}}{\mathbf{L}\mathbf{C}}$.The reactance are given by ${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{\omega L}$for inductor and ${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega C}}$for capacitor.

$\mathbf{Z}\mathbf{=}\sqrt{{\mathbf{R}}^{\mathbf{2}}\mathbf{+}{\left(X_L-X_{C)}\right)}^{\mathbf{2}}}$, where Z is the impedance

Step-2: Calculations for resonance angular frequency

${\mathrm{\omega }}_1=\frac{1}{\sqrt{\left(\mathrm{L}\right)\left(\mathrm{C}\right)}}$

Now, all the parameters are doubled

$$\begin{gathered} {\mathrm{R}}_2=2{\mathrm{R}}_1 \\ {\mathrm{C}}_2=2{\mathrm{C}}_1 \\ {\mathrm{L}}_2=2{\mathrm{L}}_1 \end{gathered}$$

Find the new resonance frequency

$$\begin{gathered} \mathrm{\omega }=\frac{1}{\sqrt{\left(2\mathrm{L}\right)\left(2\mathrm{C}\right)}} \\ =\frac{1}{\sqrt{4\mathrm{LC}}} \\ =\frac{1}{2\sqrt{\mathrm{LC}}} \\ =\frac{{\mathrm{\omega }}_1}{2} \end{gathered}$$

Therefore, resonance angular frequency is halved.

Step-3: Calculations of capacitive reactance, inductive reactance and impedance.

First calculate the inductive reactance;

$$\begin{gathered} {\left({\mathrm{X}}_{\mathrm{L}}\right)}_2={\mathrm{\omega L}}_2 \\ =\mathrm{\omega }\left(2{\mathrm{L}}_1\right) \\ =2{\left({\mathrm{X}}_{\mathrm{L}}\right)}_1 \end{gathered}$$

And then the capacitive reactance;

$$\begin{gathered} {\left({\mathrm{X}}_{\mathrm{C}}\right)}_2=\frac{1}{{\mathrm{\omega C}}_2} \\ =\frac{1}{\mathrm{\omega }\left(2{\mathrm{C}}_1\right)} \\ =\frac{{\left({\mathrm{X}}_{\mathrm{C}}\right)}_1}{2} \end{gathered}$$

Then calculate the impedance;

$$\begin{gathered} \mathrm{Z}=\sqrt{{\mathrm{R}}^2+{\left({\mathrm{X}}_{\mathrm{L}}-{\mathrm{X}}_{\mathrm{C}}\right)}^2} \\ \mathrm{Z}=\sqrt{{\left(2\mathrm{R}\right)}^2+{\left(2{\mathrm{X}}_{\mathrm{L}}-\frac{{\mathrm{X}}_{\mathrm{C})}}{2}\right)}^2} \\ \mathrm{Z}=2\sqrt{{\left(\mathrm{R}\right)}^2+{\left({\mathrm{X}}_{\mathrm{L}}-\frac{{\mathrm{X}}_{\mathrm{C})}}{4}\right)}^2} \end{gathered}$$

Therefore, a) resonance angular frequency is halved, b) the inductive reactance doubles, c) capacitive reactance halves and d) Z increases without doubling.