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.
In a 120V circuit with an open fuse, you would expect the voltage across the open fuse to be approximately 120 volts. This is because the open fuse creates a break in the circuit, preventing current from flowing, but the voltage remains present across the open points. The voltage is effectively the same as the supply voltage since there is no current to drop the voltage across the fuse.
Well, you should really measure the open-circuit voltage and the short circuit current both under dark and light conditions and then compare them to fully characterize a solar cell. Measuring the open-circuit voltage means measuring the voltage across the cell when no current is flowing (i.e., with a LARGE resistance as a load on the cell). Measuring the short-circuit current means measuring the current when the voltage across the circuit is essentially zero (i.e., with a VERY SMALL resistance as a load on the cell--thus, "short-circuit" current).
It isn't. If you're using superposition, you open circuit current sources and short voltage sources; this is because the current source declares the current that will be flowing through that branch. Both current and voltage sources have a finite internal resistance.
voltage drop is zero bcz in open ckt current will be zero
When the switch is open, the voltmeter measures the potential difference (voltage) across the terminals of the circuit components or power source it is connected to. This measurement indicates the voltage available in the circuit without any current flowing, allowing for the assessment of the electrical potential that could drive current if the circuit were closed. The reading reflects the circuit's voltage characteristics under open-circuit conditions.
Voltage is potential energy and can exist in a open circuit.
The terminal voltage is equal to the supply voltage and there is zero current.
Such a circuit either has no voltage source, or some part of the circuit is open, e.g., an open switch.
In a 120V circuit with an open fuse, you would expect the voltage across the open fuse to be approximately 120 volts. This is because the open fuse creates a break in the circuit, preventing current from flowing, but the voltage remains present across the open points. The voltage is effectively the same as the supply voltage since there is no current to drop the voltage across the fuse.
Voltage is a property of electrical potential. Amperes (and miliamperes) are the units of electrical current. Even though these are related to each other in a circuit, they are not the same thing, and they cannot be "converted" into each other.Also, these properties are only related through a "load" the circuit provides (the resistance and inductance of the circuit), and make sense only when related to each other this way. If there is current, there will be voltage as well, but if there's only voltage, there will be no current unless there is some resistance as well (even a wire has resistance) - otherwise the circuit is "open" and no charge is flowing.In a simple circuit with a voltage source and resistor:milliamps = voltage*1000/resistance.If your circuit has diodes, capacitors, inductors, etc. it gets much more complicated.
Well, you should really measure the open-circuit voltage and the short circuit current both under dark and light conditions and then compare them to fully characterize a solar cell. Measuring the open-circuit voltage means measuring the voltage across the cell when no current is flowing (i.e., with a LARGE resistance as a load on the cell). Measuring the short-circuit current means measuring the current when the voltage across the circuit is essentially zero (i.e., with a VERY SMALL resistance as a load on the cell--thus, "short-circuit" current).
It isn't. If you're using superposition, you open circuit current sources and short voltage sources; this is because the current source declares the current that will be flowing through that branch. Both current and voltage sources have a finite internal resistance.
When the switch is open, the voltmeter measures the potential difference or voltage across the two points connected by the switch, as no current flows through the circuit. This measured voltage is often referred to as the open-circuit voltage.
If the resistance is large enough, then there might not be enough voltage difference to allow much current. Since, Voltage = Current * Resistance, if resistance goes really large, and your voltage doesn't change, your current must decrease. An open circuit is where you do not have any current flowing, so whether no current verses very little current is the same is up to you.
because an ideal current source is assumed to produce a constant current for any voltage and is assumed to have an impedence of infinity (open circuit).
voltage drop is zero bcz in open ckt current will be zero
Open circuit means the circuit is not continuous . A short circuit is continuous but has a fault connecting between either live to neutral or earth .As result of this we saw that this answer is unsufficent to explain short and open circuit on the other hand you can use this answer also like i did:)