Without knowing the signals driving such a circuit and the tuning of the tank in the circuit it in not possible to give an exact answer. However assuming the tank is tuned to RF the behavior of the tank and LED should be effectively independent: the tank will respond to the RF in the signal and the LED will respond to the DC in the signal. The behavior would be different if the tank was tuned to low audio frequencies.
Chat with our AI personalities
If a DC supply is connected to the incomer of a transformer, you effectively have a short circuit, because the DC impedance of a transformer (actually, any inductor) is quite low. You will blow something.
The initial condition is the voltage and/or current existing at the time a mathematical solution begins. Example: what happens when a resistor is connected across a capacitor? well, you say that at t=0 the resistor is connected, then after that the voltage across the capacitor is v0.exp(-t/RC), where v0 is the starting voltage, t is the time, R is the resistance and C is the capacitance. This simple solution needs only one initial condition which is the starting voltage v0 across the capacitor. Linear differential equations are common in electrical engineering and a complete solution of one (such as the example) always requires one or more initial conditions.
An inductor looks like a piece of wire to DC. It will thus look like a resistor, and inductor properties do not apply.
The lamps will get dimmer. In a parallel circuit, voltage is constant. Whereas, in a series circuit, amps are constant.
When you put a light bulb in series with a inductor, the inductive reactance of the inductor reduces the current available to the light bulb, making it less bright. For this effect to be noticed, however, you need a very large inductor. To cut the current in a 60W bulb at 120VAC/60Hz by one half, for instance, you need an inductor around 0.6 henrys.