Current source means current generator for a circuit.
An ideal current source gives all current to the circuit, but practically a current source does n't give all current to the circuit, instead, a source resistor is connected in parallel to the current source to indicate the current drop.
Ideal sources do not exist in reality, they are simply theoretical entities to aid in the process of circuit analysis and design. In actual usage they are used as parts of models of real sources.
In a first approximation of a real source we must add internal resistance to the ideal source (at higher levels of approximation other things like parasitic reactances must be added to, but internal resistance is good enough for the first approximation):
for the same reason that internal resistance of voltage source is in series. the internal resistance is modeling a nonideal characteristic of the source.
in the current source the nonideal internal resistance reduces current delivery to load, so it must be in parallel to bypass some of the current around the load. in the voltage source the nonideal internal resistance blocks voltage delivery to load, so it must be in series to drop some of the voltage before the load.
This question doesn't really make sense.
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.
When a voltage source, such as a battery or a generator, is on open circuit -in other words, when it is not supplying a load- the voltage appearing across its terminals is called its 'open circuit voltage' and corresponds numerically to its electromotive force.However, when the voltage source supplies current to a load, that current also passes through the voltage source itself. This causes an internal voltage drop, which is the product of this current and the voltage source's internal resistance. This voltage drop acts in the opposite direction to the electromotive force and reduces the source's terminal voltage. This internal voltage drop will increase, of course, if either the load current increases or the internal resistance increases.So, in order to keep that the source's internal voltage drop is as low as possible, its internal resistance must be as low as possible. In the case of a battery, the internal resistance is due to the ionic resistance of the electrolyte/plates, whereas in a generator it is due to the resistance of the windings.
unit of internal resistasnce is ohms too. V = I(R+r) V voltage across the circuit I current in the circuit R external resistance r internal resistance unit of internal resistasnce is ohms too. V = I(R+r) V voltage across the circuit I current in the circuit R external resistance r internal resistance
Voltage = Current x Resistance giving us Current = Voltage / Resistance i.e. Voltage divided by resistance
An ideal voltage source has zero internal resistance so that the voltage stays constant with any load current. A practical voltage source should have less than 5% voltage drop at the rated load current.
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.
When a voltage source, such as a battery or a generator, is on open circuit -in other words, when it is not supplying a load- the voltage appearing across its terminals is called its 'open circuit voltage' and corresponds numerically to its electromotive force.However, when the voltage source supplies current to a load, that current also passes through the voltage source itself. This causes an internal voltage drop, which is the product of this current and the voltage source's internal resistance. This voltage drop acts in the opposite direction to the electromotive force and reduces the source's terminal voltage. This internal voltage drop will increase, of course, if either the load current increases or the internal resistance increases.So, in order to keep that the source's internal voltage drop is as low as possible, its internal resistance must be as low as possible. In the case of a battery, the internal resistance is due to the ionic resistance of the electrolyte/plates, whereas in a generator it is due to the resistance of the windings.
Voltage = (current) x (resistance) Current = (voltage)/(resistance) Resistance = (voltage)/(current)
Voltage = (current) x (resistance) Current = (voltage)/(resistance) Resistance = (voltage)/(current)
Voltage = (current) x (resistance) Current = (voltage)/(resistance) Resistance = (voltage)/(current)
Internal resistance. The ideal current source has no internal resistance in parallel with it (if it was set to supply no current it would act as an open circuit), and all the current it supplied would have to flow through its load (even if the load was an open circuit, in which case the voltage across the current source would be infinite). A real current source has the practical limitation that it must have an internal resistance in parallel with it, therefor some of the current it supplied is bypassed through that internal resistance and never reaches the load (if the load was an open circuit, then all the current supplied is bypassed and the resulting voltage drop across the internal resistance limits the voltage across the current source).
Take the internal series resistance of the voltage source and make it the internal parallel resistance of the current source. Then compute using Ohm's law the current of the current source to be equal to the maximum current the original voltage source could supply a short circuit load. Note: the two sources are equivalent.
unit of internal resistasnce is ohms too. V = I(R+r) V voltage across the circuit I current in the circuit R external resistance r internal resistance unit of internal resistasnce is ohms too. V = I(R+r) V voltage across the circuit I current in the circuit R external resistance r internal resistance
Ohm's law states that "The current is directly proportional to the applied EMF (voltage) and inversely proportional to the resistance in the circuit." <<>> if resistor exists, resistance decreases according to ohm's law, current is directly proportional to voltage and current is inversely proportional to resistance it means as current increases, voltage increases. resistance increases, current decreases so as voltage if there is no resistor, there should be no resistance except internal resistance of voltmeter and ammeter
Voltage = Current x Resistance giving us Current = Voltage / Resistance i.e. Voltage divided by resistance
An ideal voltage source has zero internal resistance so that the voltage stays constant with any load current. A practical voltage source should have less than 5% voltage drop at the rated load current.
The behaviour you are describing is, in fact, due to the internal resistance of the voltage source.When a voltage source, such as a battery or generator, is not connected to a load, its potential difference is simply the electromotive force (or 'no-load voltage') of that source. When a load is connected, a load current flows not only through the load itself, but also through the internal resistance of the source. This causes an internal voltage drop across the internal resistance, which acts in the opposite sense (i.e. in accordance with Kirchhoff's Voltage Law), or direction, to the electromotive force, thus reducing the voltage available at the terminals. The greater the load (i.e. the lower the load resistance), the greater the resulting load current, and the greater the internal voltage drop -and the lower the terminal voltage.