Question

In a series LCR circuit, the inductance L is 10 mH, Capacitance C is 1μF and resistance R is 100Ω. The frequency at which resonance occurs is

• A. 1.59 kHz
• B. 15.9 rad/s
• C. 15.9 kHz
• D. 1.59 rad/s

Answer 1

The Correct Option is A

To solve the problem of finding the frequency at which resonance occurs in a series LCR circuit, we need to use the formula for the resonant frequency in such a circuit. The resonant frequency \((f_0)\) is given by the formula:

\(f_0 = \frac{1}{2\pi\sqrt{LC}}\)

Where:

• \(L\) is the inductance in henries (H).
• \(C\) is the capacitance in farads (F).

Let's substitute the given values into the formula:

• \(L = 10 \, \text{mH} = 10 \times 10^{-3} \, \text{H}\)
• \(C = 1 \, \mu\text{F} = 1 \times 10^{-6} \, \text{F}\)

Substitute these values into the resonant frequency formula:

\(f_0 = \frac{1}{2\pi\sqrt{(10 \times 10^{-3}) \times (1 \times 10^{-6})}}\)

Calculate the denominator:

\(\sqrt{(10 \times 10^{-3}) \times (1 \times 10^{-6})} = \sqrt{10 \times 10^{-9}} = \sqrt{10} \times 10^{-4.5}\)

Now calculate:

\(f_0 = \frac{1}{2\pi \times \sqrt{10} \times 10^{-4.5}}\)

Approximate \(\sqrt{10} \approx 3.16\).

Therefore,

\(f_0 = \frac{1}{2\pi \times 3.16 \times 10^{-4.5}}\)

\(f_0 \approx \frac{1}{19.84 \times 10^{-4.5}}\)

Calculating further:

\(f_0 \approx 1591.55 \, \text{Hz}\)

Rounding off gives us approximately \(1591 \, \text{Hz} \approx 1.59 \, \text{kHz}\).

Thus, the frequency at which resonance occurs in the circuit is approximately 1.59 kHz, which matches option 1.