Question

the resonant frequency for a parallel rcl circuit with a 9 pf capacitor and a 5 mh inductor is approximately ____ khz
options:
750
0.024
0.012
24

Explanation:

Step1: Recall the resonant frequency formula for RLC circuit

The resonant frequency \( f \) (in Hz) for a parallel (or series) RLC circuit is given by the formula:
\[ f = \frac{1}{2\pi\sqrt{LC}} \]
where \( L \) is the inductance in henries (H) and \( C \) is the capacitance in farads (F).

Step2: Convert units to SI units

Given:
- Capacitance \( C = 9 \, \text{pF} = 9 \times 10^{-12} \, \text{F} \) (since \( 1 \, \text{pF} = 10^{-12} \, \text{F} \))
- Inductance \( L = 5 \, \text{mH} = 5 \times 10^{-3} \, \text{H} \) (since \( 1 \, \text{mH} = 10^{-3} \, \text{H} \))

Step3: Substitute values into the formula

First, calculate \( \sqrt{LC} \):
\[ LC = (5 \times 10^{-3} \, \text{H})(9 \times 10^{-12} \, \text{F}) = 45 \times 10^{-15} \, \text{H·F} = 4.5 \times 10^{-14} \, \text{H·F} \]
\[ \sqrt{LC} = \sqrt{4.5 \times 10^{-14}} \approx 6.7082 \times 10^{-7} \, \text{s} \]

Then, calculate \( 2\pi\sqrt{LC} \):
\[ 2\pi\sqrt{LC} \approx 2 \times 3.1416 \times 6.7082 \times 10^{-7} \approx 4.212 \times 10^{-6} \, \text{s} \]

Finally, calculate \( f \):
\[ f = \frac{1}{4.212 \times 10^{-6}} \approx 237400 \, \text{Hz} \approx 237.4 \, \text{kHz} \]

Wait, maybe I made a mistake in unit conversion. Let's recheck:

Wait, \( L = 5 \, \text{mH} = 5 \times 10^{-3} \, \text{H} \), \( C = 9 \, \text{pF} = 9 \times 10^{-12} \, \text{F} \).

\[ LC = (5 \times 10^{-3})(9 \times 10^{-12}) = 45 \times 10^{-15} = 4.5 \times 10^{-14} \]
\[ \sqrt{LC} = \sqrt{4.5 \times 10^{-14}} = \sqrt{45} \times 10^{-7} \approx 6.7082 \times 10^{-7} \]
\[ 2\pi\sqrt{LC} \approx 2 \times 3.1416 \times 6.7082 \times 10^{-7} \approx 4.212 \times 10^{-6} \]
\[ f = \frac{1}{4.212 \times 10^{-6}} \approx 237400 \, \text{Hz} \approx 237.4 \, \text{kHz} \]. But the options include 24 kHz? Wait, maybe I messed up the exponents. Wait, \( 5 \, \text{mH} = 5 \times 10^{-3} \, \text{H} \), \( 9 \, \text{pF} = 9 \times 10^{-12} \, \text{F} \). Let's recalculate \( LC \):

\( LC = 5 \times 10^{-3} \times 9 \times 10^{-12} = 45 \times 10^{-15} = 4.5 \times 10^{-14} \). Then \( \sqrt{LC} = \sqrt{4.5 \times 10^{-14}} = \sqrt{4.5} \times 10^{-7} \approx 2.1213 \times 10^{-7} \)? Wait, no: \( \sqrt{4.5 \times 10^{-14}} = \sqrt{4.5} \times \sqrt{10^{-14}} = \sqrt{4.5} \times 10^{-7} \approx 2.1213 \times 10^{-7} \)? Wait, no, \( \sqrt{4.5} \approx 2.1213 \), so \( 2.1213 \times 10^{-7} \). Then \( 2\pi\sqrt{LC} \approx 2 \times 3.1416 \times 2.1213 \times 10^{-7} \approx 1.333 \times 10^{-6} \). Then \( f = \frac{1}{1.333 \times 10^{-6}} \approx 750000 \, \text{Hz} = 750 \, \text{kHz} \). Wait, that's one of the options (750). Wait, where did I go wrong earlier?

Ah! Wait, \( \sqrt{4.5 \times 10^{-14}} \): \( 4.5 \times 10^{-14} = 45 \times 10^{-15} \), so \( \sqrt{45 \times 10^{-15}} = \sqrt{45} \times \sqrt{10^{-15}} = \sqrt{45} \times 10^{-7.5} \)? No, \( 10^{-14} = 10^{-7 \times 2} \), so \( \sqrt{10^{-14}} = 10^{-7} \). So \( \sqrt{4.5 \times 10^{-14}} = \sqrt{4.5} \times 10^{-7} \approx 2.1213 \times 10^{-7} \)? No, wait, \( 4.5 \times 10^{-14} = 4.5 \times 10^{-14} \), so square root is \( \sqrt{4.5} \times 10^{-7} \approx 2.1213 \times 10^{-7} \). Then \( 2\pi \times 2.1213 \times 10^{-7} \approx 1.333 \times 10^{-6} \). Then \( 1 / (1.333 \times 10^{-6}) = 750000 \, \text{Hz} = 750 \, \text{kHz} \). Yes, that matches the option 750. I must have miscalculated \( \sqrt{4.5} \) earlier. So the correct calculation is:

\[ f = \frac{1}{2\pi\sqrt{LC}} \]
\[ L = 5 \, \text{mH} = 5 \times 10^{-3} \, \text{H} \]
\[ C = 9 \, \text{pF} = 9 \times 10^{-12} \, \text{F} \]
\[ LC = (5 \times 10^{-3})(9 \times 10^{-12}) = 45 \times 10^{-15} = 4.5 \times 10^{-14} \]
\[ \sqrt{LC} = \sqrt{4.5 \times 10^{-14}} \approx 2.1213 \times 10^{-7} \, \text{s} \]
\[ 2\pi\sqrt{LC} \approx 2 \times 3.1416 \times 2.1213 \times 10^{-7} \approx 1.333 \times 10^{-6} \, \text{s} \]
\[ f = \frac{1}{1.333 \times 10^{-6}} \approx 750000 \, \text{Hz} = 750 \, \text{kHz} \]

Answer:

750