Question

L, C and R are connected in series with an alternating current source. When there will be resonance in the circuit? What will be the nature of impedance during resonance?
OR
State the Faraday's laws of electromagnetic induction. Define self induction, state the Lenz's law regarding direction of self induced current. In a coil, due to change of current at the rate of 1 A/s an e.m.f. of 1 V is produced. What will be the coefficient of self inductance of the coil?

Hint:

Series resonance occurs when \(X_L=X_C\), i.e. \(\omega_0=\dfrac{1}{\sqrt{LC}}\), where the impedance falls to its minimum value \(Z=R\) (purely resistive). For the coil, use \(\varepsilon=L\dfrac{dI}{dt}\) to find \(L\).

Answer 1

Option 1: Resonance by the phase-angle (phasor) approach

Step 1: In a series \(LCR\) circuit the voltage across the inductor leads the current by \(90^{\circ}\) while the voltage across the capacitor lags it by \(90^{\circ}\); these two are exactly opposite in phase. The phase angle \(\phi\) between the applied voltage and the current obeys \(\tan\phi=\dfrac{X_L-X_C}{R}=\dfrac{\omega L-\dfrac{1}{\omega C}}{R}\).
Step 2: The circuit is at resonance when the applied voltage and the current are exactly in phase, i.e. \(\phi=0\) so that \(\tan\phi=0\). This requires the numerator to vanish: \(\omega L-\dfrac{1}{\omega C}=0\).
Step 3: Therefore \(\omega L=\dfrac{1}{\omega C}\), which gives \(\omega_0=\dfrac{1}{\sqrt{LC}}\) and the resonant frequency \(f_0=\dfrac{1}{2\pi\sqrt{LC}}\).
Step 4: With \(X_L=X_C\) the inductor and capacitor voltages, being equal and opposite, cancel in the phasor sum, so the whole applied voltage appears across \(R\) alone. The impedance \(Z=V/I\) then reduces to just \(R\) -- its minimum, purely ohmic value. Consequently the current peaks at \(I_{max}=V/R\) and the power factor is unity.
\[\boxed{f_0=\dfrac{1}{2\pi\sqrt{LC}},\quad Z_{res}=R}\]

Option 2: Flux-linkage view of induction and the numerical

Faraday's laws (restated): (i) An e.m.f. appears in a circuit only while the magnetic flux threading it is changing. (ii) The induced e.m.f. is numerically equal to how fast that flux linkage changes, \(\varepsilon=-\dfrac{d(N\Phi)}{dt}\).

Self induction (energy/flux view): A coil carrying current \(I\) links a flux \(N\Phi\) that is directly proportional to its own current, so we may write \(N\Phi=L\,I\). Here \(L\) measures how strongly the coil links flux per unit current. Any change in \(I\) changes this self-flux and, by Faraday's second law, induces a back e.m.f. \(\varepsilon=-L\dfrac{dI}{dt}\) in the coil itself. This opposition is the origin of self induction and stores energy \(\tfrac12 L I^{2}\) in the coil.

Lenz's law: The induced current is directed so as to oppose the change producing it; on a rising current it flows so as to reduce the increase, and on a falling current it flows so as to sustain it.

Numerical (direct substitution):
Step 1: The self inductance is defined by \(L=\dfrac{|\varepsilon|}{|dI/dt|}\).
Step 2: Here \(|\varepsilon|=1\ \text{V}\) and \(\left|\dfrac{dI}{dt}\right|=1\ \text{A/s}\).
Step 3: Hence \(L=\dfrac{1}{1}=1\ \text{H}\) (one weber-turn of flux linkage is produced per ampere of current).
\[\boxed{L=1\ \text{H}}\]