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.
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.
both of these components connected in parallel will cause an oscillation of energy, meaning the capacitor will charge and then discharge through the inductor , which will then build up a magnetic field and discharge through the cap again , and this oscillation will go on for quite some time and then finally die out , and also if a multimeter is place across the cap it should short out , because it's as if you have connected a piece of wire right across it ...
An inductor can be used, in principle, but it has to be the right inductance, it will waste more power than a capacitor, and the motor would rotate the opposite way.
If you apply a higher voltage to a capacitor than it is rated it could over heat and explode.
As the capacitor charges the led will dim until it's minimum operational voltage is reached and it goes out.
is it ? are you sure ? but i know so many circuit where capacitor is connected with ac supply . still , if u connected a capacitor to dc supply , then : 1. if it is in SERIES with the dc supply , it will block all the dc current as capacitor provides infinite resistance to dc current . application : where u want to block dc current.(simple high pass filter) 2. if it is in PARALLEL with the dc supply , it will not block dc current , but if any ac current comes out from the supply , the ac current will go through the capacitor , as capacitor provides small resistance to ac current. application : a) where u want to block ac current.(simple low pass filter) b) to filter the noise (ac components) of dc supply.
infact what happens in practical capacitor action that air in between parallel plates of that capacitor it acts as a dielectric medium and leakage current starts flowing, that leakage curent is then being bypassed through a conductance in parallel with the capacitor.But in the ideal case that leakage current can't flow due to infinite resistance which can not be provided practically.
Nothing
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 supply voltage in a parallel circuit remains the same regardless of the number of additional resistors connected. The voltage across each resistor in a parallel circuit is the same as the supply voltage. Adding more resistors in parallel will increase the total current drawn from the supply.
The lamps will get dimmer. In a parallel circuit, voltage is constant. Whereas, in a series circuit, amps are constant.