Can a current source have a voltage across it?
A current source can certainly have a voltage across it. If the voltage across a current source is zero, then it is not delivering or absorbing any power. However, if the voltage
voltage
Yes, because I is a function of V, as long as we''re talking about resistors. Power is linearly proportional to voltage, though, if you''re talking about a constant current device.
Voltage drop across a single resistor and across two resistors
An intuitive way to look at is that all the voltage is dropped across two resistors, and since the resistors are the same, the voltage drop across each will be the same, each taking half.
How to calculate voltage drop over and power loss in wires
How do I calculate the voltage drop over wires given a supply voltage and a current? How do I anticipate on voltage drop so that the final load has the correct supply voltage? What will be
How much voltage/current is "dangerous"?
6 It''s not the voltage but the current that kills, is a popular yet still incorrect incomplete answer. It is the ENERGY that kills. With static electricity you will will be exposed to voltages much,
Three phase power supply
2 Line to line voltage for a 3phase network (120deg separation) is sqrt (3)*phase voltage. So for a 230V 3ph network the line-line is 400V
voltage
Voltage of "local ground" The absolute charge on local ground is not actually a thing. Voltage is only ever defined as a difference between two points, so there is no such
How are current and voltage related to torque and speed of a
Voltage instead "regulates" how fast a motor can run: the maximum speed a motor can reach is the speed at which the motor generates a voltage (named "Counter-electromotive
voltage
Voltage has exactly the same problem: one terminal can only "have a voltage" when compared to another terminal. Voltage acts like distance: voltage and distance are double
What is "forward" and "reverse" voltage when working with diodes?
The reverse voltage is the voltage drop across the diode if the voltage at the cathode is more positive than the voltage at the anode (if you connect + to the cathode). This