inversily propsanational
One trace of the oscilloscope traces have to be inverted to display the voltage and current with the correct phase relationship because you are probably connecting to both sides of a small resistor in series with the load. The supply side shows voltage, while the load side shows voltage drop for the load. You are showing differential voltage across the resistor in the second trace and, of course, its negative.If, however, you are using a current probe, perhaps a clamp-on ammeter, I would question if you have it connected correctly.
In a ce amplifier, an increase of base voltage causes the collector current to rise. This causes an increased voltage drop through the collector load resistor, so the collector voltage drops. With a cc amplifier the increase in current causes more voltage across the emitter load resistor, therefore the emitter voltage rises.
A: Because it is a voltage amplifier the current will inversely reflect the voltage across a resistor
1) in inductor there is generation of magnetic field due to flow of current . so there is phase difference in voltage and current . 2)in capacitor there is storage of charges. there is phase diff. 3)But in case of resistor there is no such things are happend . it is only a power dissipating element.therefor there is no phase difference between current and voltage.
Ohm's law states that voltage is resistance times current. In a resistor circuit, knowing two of voltage, current, or resistance, you can calculate the third.Actually, this applies to any circuit, be it resistor, capacitor, or inductor. Ohm's law still applies - it just gets more complex when the phase angle of current is not the same as the phase angle of voltage.
If there are only a resistor and a capacitor in the circuit, then the phase shift will indeed be between 0 and 90 degrees. When the resistor and capacitor are in series, the phase shift will be negative when the capacitor is connected to a source voltage and the resistor is the load. The phase shift will be positive when the resistor is connected to the source. The lower the values of R and C, the higher the frequency bandwidth.With the resistor and capacitor connected in series and the two parts connected to a current source, the phase shift will be negative. At high frequencies, the output voltages is lower, and the circuit appears as a very low impedance. At low frequencies, the circuit looks more like a resistor. Again, the phase shift will be between 0 and 90 degrees.CommentThe correct term is phase angle, not 'phase shift'. By definition, the phase angle is the angle by which the load current leads or lags the supply voltage. For an RC circuit, the current leads the voltage, so the phase angle is a leading phase angle.
A capacitor and a resistor has no effect on the supply voltage; however, this particular load combination will cause the load current to lead the supply voltage by some angle termed the 'phase angle'.
When they are in parallel the same voltage appear across both. The resistor carries a current of V/R, the inductor carries a current of V/(jwL). So the current in the inductor is 90 degrees behind in its phase.
POWER=VI. V=voltage I= current
Ohm's Law Voltage = Current x Resistance
There is phase to phase voltage in 3 phase system.AnswerYou don't get voltage 'phase-to-phase'; it's 'line-to-line'!
Kirchhoff's Voltage and Current Laws apply to circuits: series, parallel, series-parallel, and complex.If your circuit comprises just a single resistor, then they still apply. For example, the voltage drop across a single resistor will be equal and opposite the applied voltage (Kirchhoff's Voltage Law), and the current entering the resistor will be equal to the current leaving it (Kirchhoff's Current Law).