Question

A source of sinusoidal emf with constant voltage is connected in series with an oscillating circuit with quality factor

Q = 100

. At a certain frequency of the external voltage the heat power generated in the circuit reaches the maximum value. How much (in per cent) should this frequency be shifted to decrease the power generated

n = 2.0

times?

Accepted Answer

Power in AC circuits is given by the expression,

$P = \frac{V^{2} R}{R^{2} + {(X_{L} - X_{C})}^{2}}$ (Where $X_{C} & X_{L}$ are the capacitive and inductive reactance respectively.)
For resonance to occur the capacitive reactance must be equal to the inductive reactance,

$\Rightarrow X_{L} = X_{C}$

$\Rightarrow ω_{0} = \frac{1}{\sqrt{L C}}$ (because $X_{L} = ω L$ and $X_{C} = \frac{1}{ω C}$ ).

The power generated will decrease $n$ times when impedance increases $\sqrt{n}$ times (because $P = \frac{V^{2} R}{Z^{2}}$ ).

Since $ω$ does not affect $R$ , for this to happen,

${(X_{L} - X_{C})}^{2} = {(ω L - \frac{1}{ω C})}^{2} = (n - 1) R^{2}$ ( because $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$ ).

Solving in terms of $ω$ ,

$\Rightarrow$ $ω - \frac{ω_{0}^{2}}{ω} = \pm \sqrt{n - 1} \frac{R}{L} = \pm \sqrt{n - 1} 2 β$ ( where $ω_{0} = \frac{1}{\sqrt{L C}}$ ).

Therefore,

$ω^{2} \mp 2 \sqrt{n - 1} βω - ω_{0}^{2} = 0 (ω \mp \sqrt{n - 1} β)^{2} = ω_{0}^{2} + (n - 1) β^{2}$

$\Rightarrow$ $\frac{ω}{ω_{0}} = \sqrt{\frac{1 + (n - 1) β^{2}}{ω_{0}^{2}}} \pm \frac{\sqrt{n - 1} β}{ω_{0}}$ .

(Ignore the negative root because angular frequencies are never negative.)

Now,
$Q = \frac{ω}{2 β} = \frac{1}{2} \sqrt{{(\frac{ω_{0}}{β})}^{2} - 1}$

$\frac{ω_{0}}{β} = \sqrt{1 + 4 Q^{2}}$ .

Therefore, $\frac{ω}{ω_{0}} = \sqrt{1 + \frac{n - 1}{(1 + 4 Q^{2})}} \pm \frac{\sqrt{n - 1}}{\sqrt{1 + 4 Q^{2}}}$ .

For large values of $Q$ ,
$|\frac{ω - ω_{0}}{ω_{0}}| = \frac{\sqrt{n - 1}}{2 Q} = \frac{\sqrt{n - 1}}{2 Q} \times 100 % = 0.5 %$ .

I E Irodov Solutions for Chapter: OSCILLATIONS AND WAVES, Exercise 2: ELECTRIC OSCILLATIONS

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