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Voltage source inverters use the dc voltage (e.g a capacitor in parallel) as a source while the current source inverer (inductor in series) use the dc current as a source. Please note that voltage can not be changed abruptly in capacitor as current can not be changed abruptly in inductor.
What else can I help you with?
How will you convert a open current circuit to a open voltage circuit?
The first thing you need to know is the internal resistance of the current source, the voltage source will have the same internal resistance. Then compute the open circuit voltage of the current source, this will be the voltage of the voltage source. You are now done.
How do you change current into voltage?
Compute the open load voltage of the current source across its shunt resistance.This voltage becomes the voltage source's voltage.Move the current source's shunt resistance to the voltage source's series resistance.Insert the new voltage source into the original circuit in place of the current source.
If the resistance increases what will happen to voltage and current?
If you have a simple circuit. For eg: One voltage source and one resistor, then the voltage of the circuit will always remain the same, the current however will decrease following Ohms' Law V=I*R. If we have a current source instead of a voltage source, we are forcing the current to be a certain value so if we increase the resistor value the current will remain the same but the voltage will increase.
What happens to current in a circuit if the voltage is halved and the resistance stays the same?
the current doubles.. explanation:V=IR hence I=V/R which means that when the supply voltage is constant ,current is inversely proportional to resistance.thus the current doubles. practically speaking when the resistance of the load(fan ,bulb,refrigerator,....) is less ,it draws more current from the source so as to balance the voltage across it.i.e; to maintain the voltage across it as constant. This answer is absolutely correct if you assume that the current comes from a pure voltage source ( voltage source with zero internal resistance). At the other extreme you could have a current source (such as a very large voltage source in series with a very large resistor), and then the current is practically independent of changes if the external resistance is changed (because the change represents a relatively minute change in the overall resistance). With appropriate circuitry it is possible to devise a situation where the current is practically independent of the changing resistance.
Why do you short voltage source and open current source while using superposition theorem?
You short a voltage source when doing this analysis because you do not know how much current will flow through the voltage source - consider it an undefined value. For the same reason, you open a current source since you know how much current will be flowing through it. This is a simple explanation; I'm sure a more exhaustive, technical one could be made if this is not sufficient.
How will you convert a open current circuit to a open voltage circuit?
The first thing you need to know is the internal resistance of the current source, the voltage source will have the same internal resistance. Then compute the open circuit voltage of the current source, this will be the voltage of the voltage source. You are now done.
How do you change current into voltage?
Compute the open load voltage of the current source across its shunt resistance.This voltage becomes the voltage source's voltage.Move the current source's shunt resistance to the voltage source's series resistance.Insert the new voltage source into the original circuit in place of the current source.
If the resistance increases what will happen to voltage and current?
If you have a simple circuit. For eg: One voltage source and one resistor, then the voltage of the circuit will always remain the same, the current however will decrease following Ohms' Law V=I*R. If we have a current source instead of a voltage source, we are forcing the current to be a certain value so if we increase the resistor value the current will remain the same but the voltage will increase.
What happens to current in a circuit if the voltage is halved and the resistance stays the same?
the current doubles.. explanation:V=IR hence I=V/R which means that when the supply voltage is constant ,current is inversely proportional to resistance.thus the current doubles. practically speaking when the resistance of the load(fan ,bulb,refrigerator,....) is less ,it draws more current from the source so as to balance the voltage across it.i.e; to maintain the voltage across it as constant. This answer is absolutely correct if you assume that the current comes from a pure voltage source ( voltage source with zero internal resistance). At the other extreme you could have a current source (such as a very large voltage source in series with a very large resistor), and then the current is practically independent of changes if the external resistance is changed (because the change represents a relatively minute change in the overall resistance). With appropriate circuitry it is possible to devise a situation where the current is practically independent of the changing resistance.
If the voltage is increased and current remains the same, what happens to the resistance?
According to ohms law (R=V/I) if voltage increases the resistance also increases .For example: If voltage (V) becomes 2 times the resistance (R) also increases becomes 2 times keeping the current (I) same
Why do you short voltage source and open current source while using superposition theorem?
You short a voltage source when doing this analysis because you do not know how much current will flow through the voltage source - consider it an undefined value. For the same reason, you open a current source since you know how much current will be flowing through it. This is a simple explanation; I'm sure a more exhaustive, technical one could be made if this is not sufficient.
Is emf current greater than current?
EMF is electromotive force. It is another name for voltage. Voltage is electric potential in joules per coulomb. Current is electric flow, in amperes. Amperes are coulombs per second. Voltage and current are not the same thing, and "emf current", or "voltage current" does not make sense.
Is a resistor and a relay the same thing?
No, they are not the same. A resistor is a current reducer and a relay is essentially a switch (using low voltage to switch high voltage on/off)
How does the concept of parallel resistors connected to the same voltage source affect the overall resistance in a circuit?
When resistors are connected in parallel to the same voltage source, the overall resistance in the circuit decreases. This is because the current has multiple paths to flow through, reducing the total resistance that the current encounters.
What happens to the current in a circuit as a capacitor charges?
What happens to the current in a circuit as a capacitor charges depends on the circuit. As a capacitor charges, the voltage drop across it increases. In a typical circuit with a constant voltage source and a resistor charging the capacitor, then the current in the circuit will decrease logarithmically over time as the capacitor charges, with the end result that the current is zero, and the voltage across the capacitor is the same as the voltage source.
What is the relationship between the size of your source and an electric current?
If the source you're talking about is an ideal voltage source, then the amount of current depends on the size of the source and the total resistance of the circuit connected to it. Ohm's Law tells us that the current, I, is directly proportional to the voltage, V, and inversely proportional to the resistance, R: I = V/R So, increasing the voltage increases the current, whereas decreasing the resistance does the same. There are practical limitations to that, however. In the real world, reducing the resistance to zero does not produce infinite current, as suggested by the formula. Infinite current is produced only by "ideal" voltage sources, which don't exist.
What happen when the electric energy source does not match the load?
I am not quite sure about the experimental setup. But in general, the power (not energy) taken out by the load and any resistors is taken out of the source. It is impossible to take out more or less power than is generated by the source. The way this works is as follows: the load has a certain resistance; as a result, if a certain voltage is applied, a certain current will flow (use Ohm's Law to calculate the current). This same current flows through the source (Kirchhoff's Current Law), and if there are no additional resistances, the voltage will also be the same (Kirchhoff's Voltage Law). Therefore, the power will be the same.
