In a series resonant LCR circuit, the voltage across R is 100 V and R=1 k Ωwith C=2 μF. The res

step 1 In this problem first I will write the given data. So we have given v is equal to Vr which is eq

In a series resonant LCR circuit, the voltage across R is...

In a series resonant LCR circuit, the voltage across $R$ is $100 \mathrm{~V}$ and $\mathrm{R}=1 \mathrm{k} \Omega$ with $\mathrm{C}=2 \mu \mathrm{F}$. The resonant frequency 0 is $200 \mathrm{rad} / \mathrm{s}$. At resonance the voltage across $\mathrm{L}$ is. (a) $40 \mathrm{~V}$ (b) $250 \mathrm{~V}$ (c) $4 \times 10^{-3} \mathrm{~V}$ (d) $2.5 \times 10^{-2} \mathrm{~V}$

In a series resonant LCR circuit, the voltage across $R$ is $100 \mathrm{~V}$ and $\mathrm{R}=1 \mathrm{k} \Omega$ with $\mathrm{C}=2 \mu \mathrm{F}$. The resonant frequency 0 is $200 \mathrm{rad} / \mathrm{s}$. At resonance the voltage across $\mathrm{L}$ is.
(a) $40 \mathrm{~V}$
(b) $250 \mathrm{~V}$
(c) $4 \times 10^{-3} \mathrm{~V}$
(d) $2.5 \times 10^{-2} \mathrm{~V}$

Key Concepts

Series Resonant Circuit

A series resonant circuit in AC analysis consists of a resistor, an inductor, and a capacitor connected in series. At its resonant frequency, the inductive reactance and capacitive reactance are equal in magnitude but opposite in phase, causing them to cancel each other out. This results in the circuit behaving as a purely resistive load, where the total impedance is minimized and limited only by the resistor, which maximizes the current for a given applied voltage.

Resonance Condition

The resonance condition in a series LCR circuit is achieved when the angular frequency of the source, ?, equals the value given by ?? = 1/?(LC). This condition ensures that the reactive effects of the capacitor and the inductor perfectly balance each other. At resonance, although the overall impedance is minimal due to cancellation, the individual voltage drops across the reactive components can be significantly higher, illustrating the energy exchange between the magnetic field of the inductor and the electric field of the capacitor.

Voltage-Current Relationships in AC Circuits

In AC circuits, Ohm's law is extended to phasor form, where the voltage across each component is the product of the current and its impedance. For the resistor, this simply becomes V_R = I*R. However, for the reactive components like the inductor and the capacitor, the voltage is given by V_L = I*(?L) and V_C = I*(1/(?C)) respectively. At resonance, even though these reactive voltages are equal in magnitude, they are 180° out of phase, leading to their cancellation in the total circuit voltage, yet they still exist individually and can be much greater than the applied voltage.

Quality Factor (Q) and Voltage Magnification

The Quality Factor, or Q factor, in a resonant circuit is a measure of the sharpness of the resonance peak and is defined as the ratio of the reactive impedance to the resistance at resonance (Q = ?L/R for the inductor). A higher Q indicates greater voltage magnification across the reactive components relative to the applied voltage. In a series resonant circuit, the voltage across the inductor or capacitor can be Q times larger than the source voltage, demonstrating how energy is oscillated between the inductor and capacitor with minimal losses in the resistor.

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