Great! Try not to forget that the Q of the tank circuit also is a measure of the increased voltage blocking requirement of the non-polar (remember?) capacitor sitting across the inductor.

The voltage magnification that takes place at resonance is given the symbol Q and the "Q Factor" (the voltage magnification) of LC Band Pass and Band Stop filter circuits for example, controls the "rejection", the ratio of the wanted to the unwanted frequencies that can be achieved by the circuit.

The effects of voltage magnification are particularly useful as they can provide magnification of AC signal voltages using only passive components, i.e. without the need for any external power supply.

**In some cases voltage magnification can also be a ***dangerous property.* in high voltage mains (line) operated equipment containing inductance and capacitance, care must be taken during design to ensure that the circuit does not resonate at frequencies too close to that of the mains (line) supply. If that should happen, extremely high reactive voltages could be generated within the equipment, with disastrous consequences for the circuit and/or the user.

The Q factor can be calculated using a simple formula. The ratio of the supply voltage V_{S} to either of the (equal) reactive voltages V_{C} or V_{L} will be in the same ratio as the total circuit resistance (R) is to either of the reactances (X_{C} or X_{L}) at resonance. The ratio of the reactive voltage VL to the supply voltage V_{S} is the magnification factor Q.

The formula for Q is

Q = I²X / I²R … which isn't terribly helpful, but also

Q = √(L/C) / R which also simplifies to

Q = √(L/R²C); which for your component mix becomes

Q = √(10 /( 50² × 0.25×10⁻⁶ ) );

Q = 125

Which is one heck of a voltage multiplier. Putting that into perspective, the smallish A/C injection shown on your SPICE simulation diagram (20 volts, 100 Hz) is first going to be filtered by the 20 Ω front series resistor and the 220 µF capacitor to

Z = 1/(2πfC)

Z = 1/(2 × 3.1415 × 100 Hz × 220×10⁻⁶ F)

Z = 7.24 Ω

and in a voltage divider (not quite correct, but close)

V_{AC-at-L} = V_{AC} × 7.24 / (20 + 7.24);

… = 20 × 7.24 ÷ 27.24

… = 5.3 V, and now multiplied by Q

V_{PEAK} = 5.3 × 126

V_{PEAK} = 667 V

See what I mean? At resonance, the internal voltage can be REALLY large. This is the same principle used in building radio antennae. Basically, putting a capacitor in parallel of the 'helpful' value, causes the antenna to sympathetically oscillate. The resulting voltage measured across the resonant antenna can be many dozes of times higher than the actual RF voltage.

Just Saying,

**Goat**Guy ✓