Question 245: AC fundamentals

Question

$ 	ext { The frequency at resonance for the circuit is } $
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Accepted Answer

The inductors in the circuit are connected in parallel.
When we have two inductors, $L_1$ and $L_2$, in parallel, the total inductance ($L_T$) is found using the formula: $\frac{1}{L_T} = \frac{1}{L_1} + \frac{1}{L_2}$
In this circuit, both inductors are the same ($L_1 = L_2 = L$). So, $\frac{1}{L_T} = \frac{1}{L} + \frac{1}{L} = \frac{2}{L}$
This means $L_T = \frac{L}{2}$
The capacitors are also connected in parallel.
For two capacitors, $C_1$ and $C_2$, in parallel, the total capacitance ($C_T$) is: $C_T = C_1 + C_2$
In this circuit, both capacitors are the same ($C_1 = C_2 = C$), so $C_T = C + C = 2C$
But the explanation used the formula for capacitors in series, so this may be an important point to check when solving problems. If the capacitors were in series, the formula would be $\frac{1}{C_T} = \frac{1}{C_1} + \frac{1}{C_2}$ and $C_T = \frac{C}{2}$
Now, the resonant frequency ($f$) for an LC circuit is: $f = \frac{1}{2\pi \sqrt{L_T C_T}}$
Substituting $L_T = \frac{L}{2}$ and $C_T = \frac{C}{2}$, we get: $f = \frac{1}{2\pi \sqrt{\frac{L}{2} \cdot \frac{C}{2}}}$
This simplifies to: $f = \frac{1}{\pi \sqrt{LC}}$

Answer

$\frac{1}{4 \pi \sqrt{\mathrm{LC}}}$

Answer

$\frac{1}{2 \pi \sqrt{\mathrm{LC}}}$

Answer

$\frac{1}{\pi \sqrt{\mathrm{LC}}}$

Answer

$\frac{2}{\pi \sqrt{\mathrm{LC}}}$

AC fundamentals

\text{ The frequency at resonance for the circuit is }

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