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Chapter 5: Q5E (page 294)

An RLC series circuit has a voltage source of form\({\bf{E(t) = }}{{\bf{E}}_{\bf{o}}}{\bf{cos\gamma t}}\)V, a resistor of 10 Ω, an inductor of 4 H, and a capacitor of 0.01 F. Sketch the frequency response curve for this circuit.

Short Answer

Expert verified

The local minimum and the local maximum are\({\bf{\gamma = 0,\gamma = }}\sqrt {\frac{{{\bf{175}}}}{{\bf{8}}}} \)respectively.

Step by step solution

01

Use the given conditions

The differential equation is\({\bf{L}}\frac{{{{\bf{d}}^{\bf{2}}}{\bf{q}}}}{{{\bf{d}}{{\bf{t}}^{\bf{2}}}}}{\bf{ + R}}\frac{{{\bf{dq}}}}{{{\bf{dt}}}}{\bf{ + }}\frac{{\bf{1}}}{{\bf{C}}}{\bf{q = }}{{\bf{E}}_{\bf{o}}}{\bf{cos\gamma t}}\).

The frequency response curve is:

\({\bf{M(\gamma ) = }}\frac{{\bf{1}}}{{\sqrt {{{\left( {\frac{{\bf{1}}}{{\bf{C}}}{\bf{ - L}}{{\bf{\gamma }}^{\bf{2}}}} \right)}^{\bf{2}}}{\bf{ + }}{{\bf{R}}^{\bf{2}}}{{\bf{\gamma }}^{\bf{2}}}} }}\)

Here R=10 Ω,L=4H,C=0.01 F then the equation is:

\({\bf{M(\gamma ) = }}\frac{{\bf{1}}}{{\sqrt {{{{\bf{(100 - 4}}{{\bf{\gamma }}^{\bf{2}}}{\bf{)}}}^{\bf{2}}}{\bf{ + 100}}{{\bf{\gamma }}^{\bf{2}}}} }}\)

02

Find the value of \({\bf{\gamma }}\)

The curve has its local maximum.

\(\begin{aligned}{c}{\bf{M'(\gamma ) = }}\frac{{\frac{{{\bf{ - 1}}}}{{\bf{2}}}}}{{{{{\bf{((100 - 4}}{{\bf{\gamma }}^{\bf{2}}}{{\bf{)}}^{\bf{2}}}{\bf{ + 100}}{{\bf{\gamma }}^{\bf{2}}}{\bf{)}}}^{\frac{{\bf{3}}}{{\bf{2}}}}}}}{\bf{.((100 - 4}}{{\bf{\gamma }}^{\bf{2}}}{{\bf{)}}^{\bf{2}}}{\bf{ + 100}}{{\bf{\gamma }}^{\bf{2}}}{\bf{)'}}\\{\bf{ = }}\frac{{\frac{{{\bf{ - 1}}}}{{\bf{2}}}}}{{{{{\bf{(10000 - 700}}{{\bf{\gamma }}^{\bf{2}}}{\bf{ + 16}}{{\bf{\gamma }}^{\bf{4}}}{\bf{)}}}^{\frac{{\bf{3}}}{{\bf{2}}}}}}}{\bf{.(10000 - 700}}{{\bf{\gamma }}^{\bf{2}}}{\bf{ + 16}}{{\bf{\gamma }}^{\bf{4}}}{\bf{)'}}\\{\bf{ = }}\frac{{{\bf{\gamma (175 - 8}}{{\bf{\gamma }}^{\bf{2}}}{\bf{)}}}}{{{\bf{2(25000 - 175}}{{\bf{\gamma }}^{\bf{2}}}{\bf{ + 4}}{{\bf{\gamma }}^{\bf{4}}}{{\bf{)}}^{\frac{{\bf{3}}}{{\bf{2}}}}}}}\end{aligned}\)

03

Apply initial conditions.

Now,

\(\begin{aligned}{c}{\bf{M'(}}{{\bf{\gamma }}_{\bf{o}}}{\bf{) = 0}}\\\frac{{{{\bf{\gamma }}_{\bf{o}}}{\bf{(175 - 8}}{{\bf{\gamma }}^{\bf{2}}}_{\bf{o}}{\bf{)}}}}{{{\bf{2(25000 - 175}}{{\bf{\gamma }}^{\bf{2}}}_{\bf{o}}{\bf{ + 4}}{{\bf{\gamma }}^{\bf{4}}}_{\bf{o}}{{\bf{)}}^{\frac{{\bf{3}}}{{\bf{2}}}}}}}{\bf{ = 0}}\\{{\bf{\gamma }}_{\bf{o}}}{\bf{(175 - 8}}{{\bf{\gamma }}^{\bf{2}}}_{\bf{o}}{\bf{) = 0}}\\{{\bf{\gamma }}_{\bf{o}}}{\bf{ = 0,}}{{\bf{\gamma }}_{\bf{o}}}{\bf{ = }}\sqrt {\frac{{{\bf{175}}}}{{\bf{8}}}} \end{aligned}\)

The local minimum is \({\bf{\gamma }} = 0\) and the local maximum is\({\bf{\gamma }} = \sqrt {\frac{{{\bf{175}}}}{{\bf{8}}}} \).

This is the required result.

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