LC Resonance and Filter Bandwidth

A charged capacitor stores energy in an electric field. As it discharges through an inductor, current and magnetic-field energy increase. The changing current then sustains charge flow, charging the capacitor with opposite polarity. Energy alternates between electric and magnetic stores in the ideal oscillation.

Inductance plays a role analogous to mass: it resists change in current. Capacitance relates to compliance in the spring analogy, so greater capacitance corresponds to a less stiff spring. Increasing either L or C lowers the natural frequency. Only the parallel resonance arrangement is required here.

A voltage or energy response plotted against driving frequency peaks around resonance. Losses limit the peak and broaden it. The bandwidth fB is the separation between the two frequencies at which the energy response has fallen to half its maximum.

Because stored energy is proportional to voltage squared, a half-energy point corresponds to voltage amplitude $1/\sqrt2$ of its peak, not half the peak voltage. State whether the vertical axis is voltage or energy before reading bandwidth. The current-response curve is outside the required treatment.

A radio receiver needs to reject neighbouring channels but retain the wanted signal’s bandwidth. An extremely narrow resonance can suppress information-carrying sidebands even while selecting the carrier well. Filter design therefore balances selectivity and the required information bandwidth.

Use component units carefully: mH means 10⁻³ H and nF means 10⁻⁹ F. A change in capacitance by a factor of four changes frequency by a factor of one half, not one quarter.

Try each problem before opening the worked solution. State the physical model, show substitutions with consistent units, and explain what your answer means.