Question

An L-R-C series circuit has L = 0.450 H, C =$\mathbf{2}\mathbf{.}\mathbf{50}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{F}$, and resistance R. (a) what is the angular frequency of the circuit when R = 0? (b) What value must R have togive a 5.0% decrease in angular frequency compared to the value calculated in part (a)?

Short answer

A)The angular frequency of the circuit is 298rad/s

B) The value of R is$83.8\mathrm{\Omega }$

Step-by-step solution

Concept of angular frequency

The angular frequency is given as ${\mathbf{\varpi }}_{\mathbf{0}}\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{LC}}}$were L is the inductance and C is the capacitance

Calculate the angular frequency of the circuit

In an RLC circuit, with L= 0.450 H and C =$2.50\times {10}^{-5}\mathrm{F}$, the angular frequency of the circuit when R =O. In this case we have a simple LC circuit, with angular frequency of,${\mathrm{\varpi }}_0=\frac{1}{\sqrt{\mathrm{LC}}}$Substitute the values in the above equation we have,

$$\begin{gathered} {\mathrm{\varpi }}_0=\frac{1}{\sqrt{\left(0.450\mathrm{H}\right)\left(2.50\times {10}^{-5}\mathrm{F}\right)}} \\ =298\mathrm{rad}/\mathrm{s} \end{gathered}$$

Therefore, the angular frequency of the circuit is 298rad/s

STEP 3: To find value R

The value ofR such that it gives a 5.0% decrease in angular frequency wo, that is$\frac{\mathrm{\varpi }}{{\mathrm{\varpi }}_0}=0.95$.The angular frequency of the oscillations is given by$\mathrm{\varpi }=\sqrt{\frac{1}{\mathrm{LC}}-\frac{{\mathrm{R}}^2}{4{\mathrm{L}}^2}}$Divide both sides by ${\mathrm{\varpi }}_0=\frac{1}{\sqrt{\mathrm{LC}}}$we have $\frac{\mathrm{\varpi }}{{\mathrm{\varpi }}_0}=\sqrt{\frac{\frac{1}{\mathrm{LC}}-\frac{{\mathrm{R}}^2}{4{\mathrm{L}}^2}}{1/\mathrm{LC}}}$.Now,square both sides to get,${\left(\frac{\mathrm{\varpi }}{{\mathrm{\varpi }}_0}\right)}^2=\frac{\frac{1}{\mathrm{LC}}-\frac{{\mathrm{R}}^2}{4{\mathrm{L}}^2}}{1/\mathrm{LC}}=1\frac{{\mathrm{R}}^2\mathrm{C}}{4\mathrm{L}}$, Substitute the values in the above equation we have,

$$\begin{gathered} 1-\frac{{\mathrm{R}}^2\mathrm{C}}{4\mathrm{L}}={\left(0.95\right)}^2 \\ \mathrm{R}=\sqrt{\frac{4\mathrm{L}}{\mathrm{C}}}\left(1-{\left(0.95\right)}^2\right) \\ =\sqrt{\frac{4\left(0.450\mathrm{H}\right)\left(0.0975\right)}{\left(2.50\times {10}^{-5}\mathrm{F}\right)}} \\ =83.8\mathrm{\Omega } \end{gathered}$$

Therefore, the value of R is$83.8\mathrm{\Omega }$