Animated Solution for Physics - Electromagnetic Induction and Alternating Current: In the circuit shown, L=1μH, C=1μF and R=1kΩ. They are connected in series with an AC source V=V0sinωt as shown. Which of the following options is/are correct?
Select Answer:
* Multiple Correct
Visualized Solution
LCR Circuit
Series LCR Circuit
L=1μH
C=1μF
R=1kΩ
XC Formula
XC=ωC1
ω→0 Limit
limω→0XC=∞
I≈0
I≈0 at ω≈0
Resonance Condition
In phase condition: XL=XC
ωr Derivation
ωrL=ωrC1
ωr=LC1
Independence from R
ωr is independent of R
Substituting Values
ωr=10−6×10−61
Calculating ωr
ωr=10−61=106 rad/s
Checking Option C
ωr=104 rad/s
High Frequency Limit
At ω>>106 rad/s, ω>ωr
Comparing Reactances
XL=ωL (Large)
XC=ωC1 (Small)
Inductive Behavior
XL>XC⟹Inductive Circuit
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The Sigma Insight: Alternating Current (AC) and Voltage
Solution Diagram
Analyzing the Setup
Imagine you are an electron trying to push your way through this circuit. You encounter three distinct obstacles: an inductor (L), a capacitor (C), and a resistor (R). These three components are connected in series to an alternating current (AC) voltage source, V=V0sinωt. The behavior of this circuit is entirely dictated by the frequency ω of the AC source. Let's break down how the circuit responds at different extremes of frequency to determine which of the given options are correct.
The Low-Frequency Limit
First, let's explore what happens when the frequency ω is extremely low, approaching zero (ω≈0). At this limit, the AC source behaves almost like a direct current (DC) source.
The opposition offered by the capacitor is called capacitive reactance, given by the formula:
XC=ωC1
As ω→0, the denominator becomes vanishingly small, causing the capacitive reactance XC to shoot up to infinity. Physically, a capacitor consists of two parallel plates separated by an insulator. It completely blocks the flow of steady DC current. Because the reactance is infinite, the circuit acts like an open switch, and the current I becomes nearly zero. Therefore, Option (a) is correct.
The Resonance Condition
Next, we need to find the condition where the current and voltage are perfectly in phase. In an LCR circuit, the inductor causes the voltage to lead the current by 90∘, while the capacitor causes the voltage to lag by 90∘. For the overall voltage and current to be in phase, these two opposing effects must perfectly cancel each other out. This happens at a special frequency called the resonance frequency (ωr), where the inductive reactance equals the capacitive reactance:
XL=XC
Substituting their respective formulas, we get:
ωrL=ωrC1
Solving for ωr, we find:
ωr=LC1
Notice that this formula only depends on L and C. The resistance R does not appear anywhere in this equation! The resistor only dissipates energy; it doesn't affect the natural frequency of the energy sloshing back and forth between the inductor and the capacitor. Thus, the frequency at which the current and voltage are in phase is completely independent of R. Option (b) is correct.
Let's calculate this exact resonance frequency using the given values: L=1μH=10−6 H and C=1μF=10−6 F.
ωr=10−6×10−61=10−61=106 rad/s
The current and voltage are in phase at 106 rad/s, not 104 rad/s. Therefore, Option (c) is incorrect.
The High-Frequency Limit
Finally, let's examine the circuit's behavior at very high frequencies, specifically when ω>>106 rad/s. At these frequencies, ω is much greater than the resonance frequency ωr.
Let's compare the reactances:
- The inductive reactance XL=ωL becomes extremely large because it is directly proportional to ω.
- The capacitive reactance XC=ωC1 becomes extremely small because it is inversely proportional to ω.
Since XL>>XC, the inductor completely dominates the circuit's opposition to current flow. The circuit behaves predominantly like an inductor, not a capacitor. Therefore, Option (d) is incorrect.
Final Conclusion
By systematically analyzing the limits of frequency and the conditions for resonance, we have determined that the correct statements are (a) and (b).
