The current through each resistor is equal to the voltage across it divided by its resistance for series and parallel circuits.
The circuit is incomplete, you will have no continuity, no flow of current to the load or other components. The voltage across every component in the circuit is zero. No part of the circuit stores or dissipates any energy. In short, the circuit doesn't work.
The resonance effect of the LC circuit has many important applications in signal processing and communications systems.The most common application of tank circuits is tuning radio transmitters and receivers. For example, when we tune a radio to a particular station, the LC circuits are set at resonance for that particular carrier frequency.A series resonant circuit provides voltage magnification.A parallel resonant circuit provides current magnification.A parallel resonant circuit can be used as load impedance in output circuits of RF amplifiers. Due to high impedance, the gain of amplifier is maximum at resonant frequency.Both parallel and series resonant circuits are used in induction heating.LC circuits behave as electronic resonators, which are a key component in many applications:AmplifiersOscillatorsFiltersTuners
A: Actually no. If the current can be limited it may Even oscillate and each diode will behave differently even tough they are from the same family. A reverse breakdown usually means a blown diode in a circuit the hear from a hi voltage and the hi current will surpass its power dissipation and blow short most of the time.
Most conductors and electronic devices are 'exceptions' to Ohm's Law. Ohm's Law only applies when the ratio of voltage to current remains constant for variations in voltage. Materials which behave in this way are termed 'linear' or 'ohmic'. But mostmaterials do not behave in this way, and are termed 'non-linear' or 'non-ohmic'.In simple terms, if you plot a graph of voltage against current, and the result is a curve, then that material does not obey Ohm's Law.
It can hold a very large amount of charge and release the same over along period of time. It depends in what circuit you are using it. For power supply decoupling and regulation, bigger is better. If you change the capactior value in a timing or tuned circuit, it will not behave as designed.
in a parallel circuit resistance decreases increasing the current.
I'm sure I can answer this if given more details, what circuit? current is determined by voltage and impedence(resistance) in a purely resistive circuit it is pretty basic, in alternating voltage circuits the impedence is dependent on the frequency of the voltage/current. inductors and capacitors are designed into a/c circuits to force the load to behave like a resistive circuit which boils down to how much voltage divided by how much resistance you have, 120 volts divided by 10 ohms equals 12 amps. in a parallel circuit it works pretty much the same way voltage placed on the resistor divided by its resistance equals the current flowing through that leg. more details please.
The ratio of current flow through individual branches of a parallel circuit is inversely proportional to the ratio of resistance of each branch.
They are not. They are connected differently, and the voltages and currents behave in different ways.
If all you have is a voltmeter, then resistance can only be calculated. If you know the total volts applied and current, either total current if in series ckt or current through the component if parallel type ckt, then the following formulas apply: R=Eapp÷It (series ckt) R=Eapp÷IR (parallel ckt)
Do you mean why is the voltage in a circuit lower after the light bulb than before it? If so, it's because the light bulb filament has electrical resistance. When an electrical current flows through a resistance, there is a voltage drop across the resistance (Ohm's law).More fundamentally, the light bulb is producing light, which is a form of energy. The voltage drop across the light bulb comes from the fact that electrical energy is being turned into light. If voltage didn't drop, you would be producing energy from nothing. Furthermore, if there were no voltage drop, your circuit would behave the same whether you had no light bulbs, one light bulb, or eighteen million light bulbs - something that clearly can't be the case.
"Transistor" name itself revels it transfers resistance from its input to its output (Transfer of resistance). Input resistance varies when input voltage varies, similarly output resistance varies and this leads to voltage variation at the output. Thus input to output voltage variation is called amplification. this is how transistor can be used as an amplifier. If input voltage is minimum output voltage becomes maximum i.e. its output resistance becomes maximum in common emitter configuration. Thus if no voltage is applied at the input its collector resistance becomes infinite or as if open circuit. Similarly if input current is increased output current increases and out put can behave as short circuit. This is how output current can be switched off or on using no input current or with minute input current. Unlike a digital device, the transistor is an analogue device which can be switch on/off to maximum or any gradient in between. Providing a small AC voltage to the base creates an amplified analogue of this signal across the emitter and collector.
The circuit is incomplete, you will have no continuity, no flow of current to the load or other components. The voltage across every component in the circuit is zero. No part of the circuit stores or dissipates any energy. In short, the circuit doesn't work.
Capacitors store electrical charge. Imagine we have a capacitor. At time 0 seconds we connect a DC voltage across the capacitor - immediately as the voltage is connected the capacitor is at 0 volts and the maximum current (relative to the circuit resistance) flows. At this extreme the capacitor can be treated as a short circuit, so for high frequency AC volts we should treat a capacitor as being a short circuit. As time passes the current in the circuit will go down and the voltage of the capacitor will go up - this is because as the capacitor gains more charge it gains more voltage, lowering the voltage across any resistance in the circuit consequently lowering the current in the circuit. When the capacitor is virtually full no current will flow at all and the voltage across the capacitor will equal the DC source voltage. At this extreme the capacitor can be treated as an open circuit, so for low frequency AC (allowing the capacitor to fill up before the current alternates) we can treat the capacitor as being an open circuit. Technically, it is not an open/closed circuit when it comes to AC because the capacitance will results in a signal lag or lead. However, if the frequency is low/high enough the lag/lead is often negligable.
It wouldn't light up.
If you double the the RC circuit input frequence, the magnitude of voltage and current depends on whether or not the RC circuit is configured low-pass or high-pass, and on whether or not the original frquency is close to the knee frequency (the -3db point). The question has insufficient information to be answered completely.
This depends on the circuit in question. If the circuit only has resistors and maybe incandescent light bulbs, then with an equvalent RMS voltage of AC, to the previous DC, the circuit will behave almost the same. If the circuit has components such as capacitors and inductors, then the current will be shifted to flow at a waveform which no longer matches the voltage waveform. If you're talking about a circuit which was designed to run on a 12 volt battery, then you go and plug it into the wall, it will probably break, as the equivalent voltage causes a much higher current than these components were designed to handle.