Question

At resonance, voltage across L and C in series circuit is

• A. equal and opposite
• B. unity
• C. infinite
• D. zero

Hint:

While the individual voltages $\vec{V}_L$ and $\vec{V}_C$ can be exceptionally large (magnified by the quality factor $Q$ of the circuit such that $V_L = V_C = Q \cdot V_{\text{in}}$), their combined series combination yields a total voltage drop of exactly zero: \[ \vec{V}_{LC} = \vec{V}_L + \vec{V}_C = j\vec{I}X - j\vec{I}X = 0 \] This makes a series resonant circuit act like a short circuit for the reactive elements!

Answer 1

The Correct Option is A

Concept: Consider a series RLC network driven by a sinusoidal AC voltage source of angular frequency $\omega$. The total complex impedance $Z$ of this series configuration is mathematically formulated as: \[ Z = R + j\left(X_L - X_C\right) = R + j\left(\omega L - \frac{1}{\omega C}\right) \] Where $X_L = \omega L$ is the inductive reactance and $X_C = \frac{1}{\omega C}$ is the capacitive reactance. The condition of electrical resonance is established when the inductive and capacitive reactances perfectly balance each other out, thereby rendering the imaginary part of the total input impedance zero: \[ X_L = X_C \quad \Rightarrow \quad \omega_0 L = \frac{1}{\omega_0 C} \quad \Rightarrow \quad \omega_0 = \frac{1}{\sqrt{LC}} \] At this resonant angular frequency $\omega_0$, the total impedance drops to its minimum value, which is purely resistive: $Z_0 = R$. Detailed Step-by-Step Mathematical Analysis of Voltages: Let the steady-state alternating current flowing through the series combination be represented as a phasor $\vec{I}$. Because it is a series circuit, the identical current vector $\vec{I}$ traverses all three individual elements.

Step 1: Determine the individual phasor voltage drops.
• The voltage across the pure inductor ($\vec{V}_L$) leads the current phasor by exactly $90^\circ$ ($+\frac{\pi}{2}$ radians): \[ \vec{V}_L = j \cdot \vec{I} X_L = \vec{I} \cdot X_L \angle 90^\circ \]
• The voltage across the pure capacitor ($\vec{V}_C$) lags the current phasor by exactly $90^\circ$ ($-\frac{\pi}{2}$ radians): \[ \vec{V}_C = -j \cdot \vec{I} X_C = \vec{I} \cdot X_C \angle -90^\circ \]

Step 2: Evaluate the relative properties at resonance. At resonance, we have $X_L = X_C$. Let this common value be $X$. Substituting this back into the expressions: \[ \vec{V}_L = j \vec{I} X \quad \text{and} \quad \vec{V}_C = -j \vec{I} X \] Comparing these two expressions directly: \[ \vec{V}_L = - \vec{V}_C \] This vector identity confirms that the magnitudes are precisely identical ($|\vec{V}_L| = |\vec{V}_C|$), but their phase angles are separated by exactly $180^\circ$ ($\pi$ radians). Hence, they are completely equal in magnitude and directly opposite in phase direction.