Find the peak current and resonant frequency of the following circuit (as shown in figure).
(a) 0.2

Q: Why do we subtract $X_L$ and $X_C$ in the impedance formula?

In an AC circuit, the voltage across the inductor leads the current by $90^\circ$, while the voltage across the capacitor lags by $90^\circ$. This makes them $180^\circ$ out of phase, so their reactances directly oppose each other, leading to the subtraction.

Q: Why did we calculate peak current instead of RMS current?

The question specifically asks for the 'peak current'. Peak current is calculated using the peak voltage $V_0$ ($I_0 = V_0 / Z$). If it asked for RMS current, we would have used $V_{rms} = V_0 / \sqrt{2}$.

Q: Can we use $\omega$ directly as the resonant frequency?

No, $\omega$ is the angular frequency of the applied AC source. The resonant frequency $f_0$ is an intrinsic property of the circuit, depending only on the values of $L$ and $C$, and is the frequency at which the circuit would naturally oscillate.

Q: Is $\sqrt{10} \approx \pi$ a valid approximation for JEE?

Yes, absolutely! $\pi^2 \approx 9.87$, which is very close to 10. This approximation is frequently used in physics problems to simplify complex calculations without significantly affecting the final answer.

Q: What happens if $X_L$ equals $X_C$?

If $X_L = X_C$, their effects cancel out completely. The circuit is then in a state of 'resonance', and the total impedance becomes purely resistive ($Z = R$). At this point, the current in the circuit is maximum.

Question

Find the peak current and resonant frequency of the following circuit (as shown in figure).

Options:

(A) 0.2 A and 50 Hz

(B) 0.2 A and 100 Hz

(C) 2 A and 100 Hz

(D) 2 A and 50 Hz

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome to the fascinating world of Alternating Current! In this problem, we are dealing with a classic series LCR circuit. Imagine an electrical obstacle course where the current has to navigate through an inductor ($L$), a capacitor ($C$), and a resistor ($R$), all driven by an oscillating AC voltage source.

From the given circuit diagram, we can extract our key players:
- Inductance, $L = 100	ext{ mH} = 100 	imes 10^{-3}	ext{ H}$
- Capacitance, $C = 100	ext{ }\mu	ext{F} = 100 	imes 10^{-6}	ext{ F}$
- Resistance, $R = 120	ext{ }\Omega$

The AC source is described by the equation $V = 30\sin(100t)$. By comparing this to the standard form $V = V_0\sin(\omega t)$, we immediately spot our peak voltage $V_0 = 30	ext{ V}$ and our angular frequency $\omega = 100	ext{ rad/s}$.

### The Master Equation for Impedance

To find the peak current, we need to know the total opposition the circuit offers to the AC current. This total opposition is called **Impedance** ($Z$). Unlike simple DC circuits where we just add resistances, in AC circuits, the inductor and capacitor introduce phase shifts.

The formula for impedance in a series LCR circuit is:
$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

Here, $X_L$ is the inductive reactance and $X_C$ is the capacitive reactance. Let's calculate them one by one.

For the inductor:
$$X_L = \omega L = 100 	imes (100 	imes 10^{-3}) = 10	ext{ }\Omega$$

For the capacitor:
$$X_C = \frac{1}{\omega C} = \frac{1}{100 	imes (100 	imes 10^{-6})} = 100	ext{ }\Omega$$

Now, we plug these into our impedance formula:
$$Z = \sqrt{120^2 + (10 - 100)^2}$$
$$Z = \sqrt{14400 + (-90)^2}$$
$$Z = \sqrt{14400 + 8100} = \sqrt{22500} = 150	ext{ }\Omega$$

### Calculating the Peak Current

With the total impedance $Z$ in hand, finding the peak current ($I_0$) is a breeze. We simply use the AC equivalent of Ohm's Law:
$$I_0 = \frac{V_0}{Z}$$

Substituting our values:
$$I_0 = \frac{30}{150} = 0.2	ext{ A}$$

So, the peak current flowing through the circuit is **$0.2	ext{ A}$**.

### Unveiling the Resonant Frequency

The second part of the question asks for the resonant frequency ($f_0$). Resonance occurs when the inductive reactance perfectly cancels out the capacitive reactance ($X_L = X_C$). The formula for the resonant frequency is beautifully symmetric:
$$f_0 = \frac{1}{2\pi\sqrt{LC}}$$

Let's substitute our $L$ and $C$ values:
$$f_0 = \frac{1}{2\pi\sqrt{(100 	imes 10^{-3}) 	imes (100 	imes 10^{-6})}}$$
$$f_0 = \frac{1}{2\pi\sqrt{10^{-5}}}$$

To simplify this, we can rewrite $\sqrt{10^{-5}}$ as $10^{-2.5}$, which is $\frac{\sqrt{10}}{1000}$. 
$$f_0 = \frac{100\sqrt{10}}{2\pi}$$

Here is a pro-tip for JEE physics: $\pi^2 \approx 9.87$, which is very close to $10$. Therefore, we can safely approximate $\sqrt{10} \approx \pi$. Using this approximation:
$$f_0 \approx \frac{100\pi}{2\pi} = 50	ext{ Hz}$$

### The Final Verdict

We have successfully navigated the LCR circuit! The peak current is **$0.2	ext{ A}$** and the resonant frequency is **$50	ext{ Hz}$**. This perfectly matches option (a). Keep practicing these concepts, and soon, AC circuits will feel like second nature to you!

Animated Solution for Physics - Electromagnetic Induction: Find the peak current and resonant frequency of the following circuit (as shown in figure).

Select Answer:

Visualized Solution

$L C R$ Circuit Analysis

$I_{0} = \frac{V _{0}}{Z}$

$V_{0}$ and $ω$

$X_{L}$ and $X_{C}$

$Z$

$I_{0}$

$f_{0}$

$f_{0}$

Approximation

Phase Difference

The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Master Equation for Impedance

Calculating the Peak Current

Unveiling the Resonant Frequency

The Final Verdict

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Select Answer:

Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I0​=ZV0​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } V0​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ω

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f0​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f0​

Approximation

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The Sigma Insight: Alternating Current (AC) and Voltage

Analyzing the Setup

The Master Equation for Impedance

Calculating the Peak Current

Unveiling the Resonant Frequency

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