required to form a parallel resonance filter.","back":"

Inductor ($L$)
Capacitor ($C$)

","points":1,"mcqOptions":[{"text":"

Inductor ($L$)
Capacitor ($C$)

","correct":true,"explanation":"An inductor and capacitor exchange energy between the inductor's magnetic field and the capacitor's electric field, allowing resonance at a particular frequency."},{"text":"

Resistor ($R$)
Capacitor ($C$)

","correct":false,"explanation":"A resistor-capacitor combination forms an RC filter and cannot produce LC resonance. A parallel resonance filter requires an inductor and capacitor to exchange stored energy."},{"text":"

Inductor ($L$)
Resistor ($R$)

","correct":false,"explanation":"An inductor-resistor combination forms an RL filter, with no electric-field energy store. Resonance requires both an inductor and capacitor to exchange magnetic and electric energy."},{"text":"

Transformer ($T$)
Capacitor ($C$)

","correct":false,"explanation":"A transformer may contain inductive coils, but it is not the required component in the basic resonant pair. The filter is formed from an inductor and capacitor."}]},{"id":"def1083b-095f-4a14-b99a-e9853c662e15","cardType":"basic","front":"State the formula for the resonant frequency ($f_0$) of an LC circuit.","back":"$$$f_0 = \\frac{1}{2\\pi\\sqrt{LC}}$$","points":1,"mcqOptions":[{"text":"$$$f_0 = \\frac{1}{2\\pi\\sqrt{LC}}$$","correct":true,"explanation":"At resonance, the inductive and capacitive effects balance. Solving the LC angular-frequency relation $\\omega_0=1/\\sqrt{LC}$ and using $f_0=\\omega_0/(2\\pi)$ gives this formula."},{"text":"$$$f_0 = \\frac{1}{2\\pi LC}$$","correct":false,"explanation":"This omits the square root applied to the product $LC$. The angular frequency is $1/\\sqrt{LC}$, so $f_0=1/(2\\pi\\sqrt{LC})$."},{"text":"$$$f_0 = \\frac{1}{\\pi\\sqrt{LC}}$$","correct":false,"explanation":"This misses the factor of $2$ when converting angular frequency to ordinary frequency. Since $\\omega_0=2\\pi f_0$, the denominator must be $2\\pi\\sqrt{LC}$."},{"text":"$$$f_0 = \\frac{1}{2\\pi}\\sqrt{\\frac{L}{C}}$$","correct":false,"explanation":"This incorrectly divides the component values instead of multiplying them, and it has the wrong dimensions for frequency. The correct relation is $f_0=1/(2\\pi\\sqrt{LC})$."}]},{"id":"5fd5f691-1065-4ea2-8e60-6c99bab1ac41","cardType":"basic","front":"Describe the mechanism that drives charge onto the capacitor's opposite plate during LC oscillation.","back":"The inductor's magnetic field collapses, maintaining current flow even after the capacitor has discharged.","points":2,"mcqOptions":[{"text":"The inductor's magnetic field collapses, maintaining current flow even after the capacitor has discharged.","correct":true,"explanation":"The collapsing magnetic field induces an emf that opposes the decrease in current. This sustains current long enough to charge the capacitor again with opposite polarity."},{"text":"The capacitor's electric field collapses, maintaining current flow even after the inductor has discharged.","correct":false,"explanation":"This reverses the roles of the two components: an inductor, not a capacitor, stores energy magnetically. Its collapsing magnetic field sustains current and charges the opposite plate."},{"text":"The inductor's magnetic field strengthens, reversing current flow before the capacitor has discharged.","correct":false,"explanation":"The magnetic field initially strengthens while the capacitor discharges, but it does not reverse the current then. Its later collapse sustains the current and charges the opposite plate."},{"text":"The capacitor's electric field strengthens, stopping current flow as soon as the inductor has charged.","co

LC resonance filters (A-level only)

LC resonance filters (A-level only) Flashcards

LC resonance filters (A-level only)

LC resonance filters (A-level only) Flashcards