JEE Main 2021
LEVELJEE Advanced
Visualized Solution
The Sigma Insight: Alternating Current (AC) and Voltage
Imagine you are an electrical engineer tasked with designing a heating element. You have two different circuit designs in front of you, and you need them to dissipate the exact same amount of average power. This is the beautiful puzzle we are about to solve!
The magic of AC circuits lies in their dynamic nature. Let's break down the two circuits we have.
Analyzing the Setup
When we look at Circuit A, it's a straightforward, no-nonsense setup. It is a purely resistive circuit containing only a resistor connected to a AC source.
Circuit B, however, is a different beast. It's a series LCR circuit. It contains the same resistor, but it also has a inductor and a capacitor. Both circuits are driven by the same source.
The Master Equation
To find the condition where both circuits dissipate the same power, we need to write down their power equations. For the purely resistive Circuit A, the average power is simply:
For the LCR Circuit B, the power depends on the impedance and the power factor . The power factor is the unsung hero of AC power; it tells us how much of the total power is actually doing useful work. The average power is:
Substituting and , we get:
The Resonance Revelation
The problem states a very interesting condition: the average power dissipated in one cycle is the same for both circuits. So, let's equate and :
Notice how beautifully the terms cancel out from both sides. Cross-multiplying, we find that:
This is a massive hint! We know the expression for impedance in an LCR circuit is . Substituting this into our equation:
The terms cancel out, leaving us with , which means:
This is the hallmark of resonance! The inductor and capacitor are engaged in a perfect cosmic dance, effectively canceling each other's reactance out.
Final Calculation
Since the inductive reactance equals the capacitive reactance, we can write . Rearranging this gives us the classic formula for resonant angular frequency:
Now, let's bring in the values from our circuit. We substitute and :
Let's calculate carefully. . The square root of is .
One divided by this value gives us , which is exactly:
So, at an angular frequency of , the LCR circuit hits resonance. It behaves purely resistively, making its power dissipation identical to Circuit A. What an elegant result!
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\begin{circuitikz}[scale=1.2] \begin{scope}[id=branch_battery] \draw (4,0) to[battery1, l={$20\,\mathrm{V}$}] (0,0); \end{scope} \begin{scope}[id=branch_R] \draw (0,0) to[R, l={$R_0 = 5\,\Omega$}] (0,2); \end{scope} \begin{scope}[id=branch_L] \draw (0,2) to[L, l_={$L = 25\,\mathrm{mH}$}] (4,2); \end{scope} \begin{scope}[id=branch_K1] \draw (4,2) to[nos, l={$K_1$}] (4,0); \end{scope} \begin{scope}[id=branch_top] \draw (0,2) -- (0,3.5) to[nos, l={$K_2$}] (2,3.5) to[C, l={$C_0 = 10\,\mu\mathrm{F}$}] (4,3.5) -- (4,2); \end{scope} \end{circuitikz}
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