A 100 ohm resistance, a 0.1 uF capacitor and an inductor are connected in series across a 250 V supp

Q: Why do we ignore the resistance and voltage in this calculation?

The resonant frequency of a series LCR circuit depends exclusively on the values of the inductor ($L$) and the capacitor ($C$). The resistance and the supply voltage affect the amplitude of the current at resonance, but not the frequency at which resonance occurs.

Q: What does resonance physically mean here?

At resonance, the energy oscillates perfectly back and forth between the magnetic field of the inductor and the electric field of the capacitor. The circuit behaves as if it's purely resistive, allowing maximum current to flow.

Q: Why is $\pi^2$ sometimes taken as 10?

$\pi^2$ is approximately 9.8696. In many competitive exams, taking $\pi^2 \approx 10$ is a standard approximation to speed up manual calculations without significantly altering the final answer choice.

Q: What happens to the impedance at resonance?

At resonance, the inductive reactance ($X_L$) and capacitive reactance ($X_C$) cancel each other out. The total impedance of the circuit becomes minimum and is exactly equal to the resistance ($R$).

Q: Can resonance occur in a DC circuit?

No, resonance requires an alternating source to drive the oscillations. In a DC circuit, the frequency is zero, so the concept of inductive and capacitive reactance balancing each other out does not apply.

Question

A $100\ \Omega$ resistance, a $0.1\ \mu	ext{F}$ capacitor and an inductor are connected in series across a $250\ 	ext{V}$ supply at variable frequency. Calculate the value of inductance of inductor at which resonance will occur. Given that the resonant frequency is $60\ 	ext{Hz}$.

Options:

(A) 0.70 H

(B) 70.3 mH

(C) $7.03 	imes 10^{-5}\ 	ext{H}$

(D) 70.3 H

Sigma Percentile · Accepted Answer

### The Symphony of Resonance

Imagine a playground swing. If you push it at just the right moment, the swing goes higher and higher. This perfect timing is what physicists call **resonance**. In the world of electronics, an LCR circuit has its own natural rhythm, a specific frequency where it 'swings' with maximum energy.

### Analyzing the Setup

We are given a series LCR circuit. This means our resistor ($R = 100\ \Omega$), capacitor ($C = 0.1\ \mu	ext{F}$), and an unknown inductor ($L$) are holding hands in a single line, connected to an alternating voltage source of $250\ 	ext{V}$. The problem tells us that this circuit hits its sweet spot—its resonant frequency—at exactly $60\ 	ext{Hz}$.

### The Master Equation

What exactly happens at resonance? The inductor and the capacitor are like two arm wrestlers perfectly matched in strength. The inductive reactance ($X_L$) perfectly cancels out the capacitive reactance ($X_C$). Mathematically, this beautiful balance gives birth to the resonant frequency formula:

$$f_0 = \frac{1}{2\pi\sqrt{LC}}$$

This equation is our golden key. We know $f_0$ and we know $C$. Our mission is to rescue $L$ from inside that square root.

### Isolating the Unknown

Let's perform some algebraic gymnastics. First, we square both sides to shatter the square root:

$$f_0^2 = \frac{1}{4\pi^2 L C}$$

Now, we swap $L$ and $f_0^2$ to make $L$ the subject of our equation:

$$L = \frac{1}{4\pi^2 f_0^2 C}$$

### The Final Calculation

It's time to plug in the numbers. We must be incredibly careful with units. The capacitance is given in microfarads, so $C = 0.1 	imes 10^{-6}\ 	ext{F} = 10^{-7}\ 	ext{F}$. The frequency is $60\ 	ext{Hz}$.

$$L = \frac{1}{4\pi^2 (60)^2 (10^{-7})}$$

Let's simplify the denominator. The square of $60$ is $3600$.

$$L = \frac{1}{4\pi^2 	imes 3600 	imes 10^{-7}}$$

We can flip the $10^{-7}$ to the numerator as $10^7$:

$$L = \frac{10^7}{4\pi^2 	imes 3600} = \frac{10^5}{4\pi^2 	imes 36}$$

Using the approximation $\pi^2 \approx 9.87$, the denominator becomes roughly $4 	imes 9.87 	imes 36 \approx 1421.28$.

$$L \approx \frac{100000}{1421.28} \approx 70.3\ 	ext{H}$$

And there we have it! The inductance required to make this circuit resonate at $60\ 	ext{Hz}$ is $70.3\ 	ext{H}$. This perfectly matches option (d).

Select Answer:

Visualized Solution

$LCR Circuit Setup$

$Condition for Resonance$

$Rearranging for L$

$Substituting Values$

$Simplifying Denominator$

$Final Calculation$

$Conclusion$

$Food for Thought$

The Sigma Insight: Alternating Current (AC) and Voltage

The Symphony of Resonance

Analyzing the Setup

The Master Equation

Isolating the Unknown

The Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } LCR Circuit Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Resonance

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rearranging for L

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Simplifying Denominator

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Food for Thought