In Fig. 31-33, a generator with an adjustable frequency of oscillation is connected to resistance $\mathit{R}\mathbf{=}\mathbf{100}\mathbf{}\mathbf{\text{Ω}}$, inductances ${\mathit{L}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{70}\mathbf{}\mathbf{\text{mH}}$ and ${\mathit{L}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{30}\mathbf{}\mathbf{\text{mH}}$, and capacitances ${\mathit{C}}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{μF}}$, ${\mathit{C}}_{\mathbf{2}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{μF}}$ , and ${\mathit{C}}_{\mathbf{3}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{50}\mathbf{}\mathbf{\text{μF}}$ . (a) What is the resonant frequency of the circuit? (Hint: See Problem 47 in Chapter 30.) What happens to the resonant frequency if (b) Ris increased, (c) $\mathbf{}{\mathit{L}}_{\mathbf{1}}$is increased, and (d) ${\mathit{C}}_{\mathbf{3}}$ is removed from the circuit?

1. Values of inductors,$L_1=1.70\text{mH},L_2=2.30\text{mH}$
2. Values of capacitors,$C_1=4.00\text{μF,}{C}_2=2.50\mathrm{\mu F}\text{,}{C}_3=3.50\mathrm{\mu F}$
3. Resistance value,$R=100\text{Ω}$

We have to use the condition of resonance for the series RLC circuit to find the resonant frequency. During resonance, the capacitive reactance and inductive reactance are the same. By using the concept of series and parallel combinations of inductors and capacitors, we can find the required quantities.

Formulae:

The equivalent inductance of a series combination,

${\mathit{L}}_{\mathit{e}\mathit{q}}\mathbf{=}\sum _{\mathit{i}}^{\mathit{n}}{\mathit{L}}_{\mathit{i}}$ ...(i)

The equivalent capacitance of a parallel combination,

${\mathit{C}}_{\mathit{e}\mathit{q}}\mathbf{=}\sum _{\mathit{i}}^{\mathit{n}}{\mathit{C}}_{\mathit{i}}$ ...(ii)

The resonant frequency of a LC circuit,

$\mathit{\omega }\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}\mathit{\pi }\sqrt{\mathit{L}\mathit{C}}}$ ...(iii)

We have two inductors in series.

Thus, by series combination of inductors, we have the equivalent inductance using equation (i) as follows:

$$\begin{gathered} L_{eq}=L_1+L_2 \\ =1.70\text{mH}+2.30\text{mH} \\ =4.00\text{mH} \\ =4.0\times {10}^{-3}\text{H} \end{gathered}$$

We have three capacitors in parallel.

Now thus for the parallel combination of capacitors, we have the equivalent capacitance using equation (ii) as follows:

$$\begin{gathered} C_{eq}=C_1+C_2+C_3 \\ =\left(4.00+2.50+3.50\right)\times {10}^{-6}\text{F} \\ =10\times {10}^{-6}\text{F} \end{gathered}$$

Now theresonant frequency is given using equation (iii) as follows:

$$\begin{gathered} \omega =\frac{1}{2\pi \sqrt{(\left(4.0\times {10}^{-3}\right)\text{H}\times \left(10\times {10}^{-6}\right) \text{F}})} \\ =795.77\text{Hz} \\ \approx 796\text{Hz} \end{gathered}$$

Hence, the value of the frequency is $796\text{Hz}$.

Since the resonant frequency $\omega$is independent of resistance, there is no effect of increasing resistance. Thus, resonant frequency $\omega$ does not change.

As resonance frequency $\omega$is inversely proportional to L, if $L_1$ is increased, $\omega$ decreases.

As resonance frequency $\omega$is inversely proportional to, if $C_3$ is removed, i.e., as

C decreases, $\omega$increases.