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No.
Soon after its anode current falls to zero, the device is not is a state to block the forward voltage due to the presence of carriers in its four layers its still in conduction mode. Thus at first it takes some time to remove excess charge from the four layers.
Thus just after its anode currents decays to zero it is unable to block the forward voltage.
What else can I help you with?
'why can't we determine barrier potential by using a voltmeter connected across the ends of a diode'?
The barrier potential is not a voltage created inside the diode. instead, it refers to the depleted zone around the juncture. Since this region is deplete from carriers (electrons or holes), it became a virtual isolator. In order to make the depleted zone conductive, you need to apply an external voltage to the diode terminals. If the voltage is in forward bias (+ to the anode and - to the cathode), you will need 0.2/0.3 V for germanium diodes and 0.6/07 V for silicon diodes. You need an external diode to keep the forward current with safe limits. If the voltage is in reverse mode (- to the anode and + to the cathode), you will need to apply much more voltage to achieve conduction, although this could permanently damage the diode. Zenner diodes, for instance, always work in reverse bias to create a stable voltage, which is used for regulation purposes.
What is external resistance?
External voltage is the ration when there is an increase in current and voltage. If you apply voltage to the outside of a circuit and need to figure out the amount of current flow, that would be the external resistance.
Kirchoff's Voltage and Current Laws apply to all ac circuits?
Kirchoff's Voltage and Current Laws apply to all AC circuits as well as DC circuits. Other laws, such as Ohm's law and Norton and Thevanin equivalents apply equally as well. The complicating factor is that, at AC, current and voltage are not usually in phase with each other, unless it is a simple resistive circuit. That makes the math harder, but it does not make it invalid or impossible.
What is knee point voltage of current transformer?
10 % increase in voltage gives you 50 % increase in excitation current is called knee point voltage. To measure this first demagnetise the CT and apply voltage gradually from secondary keeping primary winding open circuited. while doing this above phenomeneo will be obsesrved.
How could you calculate the resistance of a circuit?
A variety of techniques can be used. Node-Voltage and Mesh-current (or Loop-current) methods, for example. See related link for examples. If there is a single voltage source in the model, then find the current supplied, and Resistance = Voltage/Current. {R = V/I} You could simulate/ model the circuit on a computer then apply the power profile and a current value will be calculated. Or if you have the circuit working, Place an amp meter into the circuit and measure the current. V/i = resistance.
How do you make a LED light?
A: Most LED needs a minimum voltage and minimum current in the right direction positive anode negative cathode to light up. Unfortunately the voltage can be as low as 1.2 to 5v and the same apply to the current from 5ma to 500ma. So there is no set values it depends on the device.
How do you make LED lighting?
A: Most LED needs a minimum voltage and minimum current in the right direction positive anode negative cathode to light up. Unfortunately the voltage can be as low as 1.2 to 5v and the same apply to the current from 5ma to 500ma. So there is no set values it depends on the device.
Why does a plateau exist in the Geiger-Muller region?
In a Geiger-Müller (GM) tube, there is a central anode and a "case" that is the cathode. A voltage is applied across these two elements, and an ionizing particle passing through the GM tube will cause current flow. But how much? Let's step through things and check it out. At low voltage, any electrons released by the cathode will eventually be collected by the anode, but there is no appreciable "current" per se in this, the ionization region. Things are still pretty "tame" in the GM tube through this range of voltages. By applying more voltage, an ionizing event will generate more current flow, and this current flow will be proportional to the voltage in what is (naturally) the proportional region. And as we apply more voltage, gas amplification, or Townsend avalanche, which appeared at the beginning of this region, is increasing across the area of the anode. As we apply even more voltage, it will only make for limited additional current flow in an ionizing event because the limits of the geometry of the GM tube and of the gas media to ionize and "conduct more" with the increasing voltage are being reached. This is the limited-proportional region. As voltage is increased even more, we enter the Geiger-Müller region. In this region, the current avalanche in an ionizing event is so great that is causes a "shield" of positive ions around the anode. The high current "sucks up" all the electrons and blankets the anode in a positive field that prevents additional current flow even with an increase in voltage. This is the Geiger plateau. It's the operating region where additional differential voltage will not cause higher current flow in an ionizing event.
'why can't we determine barrier potential by using a voltmeter connected across the ends of a diode'?
The barrier potential is not a voltage created inside the diode. instead, it refers to the depleted zone around the juncture. Since this region is deplete from carriers (electrons or holes), it became a virtual isolator. In order to make the depleted zone conductive, you need to apply an external voltage to the diode terminals. If the voltage is in forward bias (+ to the anode and - to the cathode), you will need 0.2/0.3 V for germanium diodes and 0.6/07 V for silicon diodes. You need an external diode to keep the forward current with safe limits. If the voltage is in reverse mode (- to the anode and + to the cathode), you will need to apply much more voltage to achieve conduction, although this could permanently damage the diode. Zenner diodes, for instance, always work in reverse bias to create a stable voltage, which is used for regulation purposes.
Why is there a difference in current in a coil when you apply dc voltage and then AC voltage to it?
A coil has both resistance and inductance. When you apply a d.c. voltage, the opposition to current is the resistance of the coil. When you apply an a.c. voltage, the opposition to current is impedance -the vector-sum of the coil's resistance and its inductive reactance. Inductive reactance is proportional to the inductance of the coil and the frequency of the supply.
What is external resistance?
External voltage is the ration when there is an increase in current and voltage. If you apply voltage to the outside of a circuit and need to figure out the amount of current flow, that would be the external resistance.
Does Kirchhoff current law and Kirchhoff voltage law depend on the relationship between current and voltage in a resistor?
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).
How do we usually think of voltage as the cause and currents the effect?
You apply a voltage across a load and the result is that a current flows through the load. So you must have the voltage present, the cause, before current flow, the effect. Think of voltage as pressure and current as flow.
What is the difference in the current and voltage?
Current (amperes) is the rate of flow of electric charge, in coulombs per second. Voltage, on the other hand, is the electric potential of that charge, in joules per coulomb.Current and voltage are related to resistance by Ohm's Law, which states that voltage is equal to current times resistance.There is a tendency to misuse the term "current", and to apply it, for instance as "an electric current of 120 volts". This usage is incorrect. Current is current, and voltage is voltage, as noted above.
How electrical resistance can be measured in copper and aluminium?
Same as for any other substance: Apply a voltage and measure the current flow. (Resistance equals voltage divided by the current flow.)
How are current and voltage related to the power supplied by the source?
in ac circuits power,P=VICOS@ @ is the angle between voltage and current. in dc P=VI V is the voltage I is the current. Power (in Watts) is current (A) x voltage (V)
Kirchoff's Voltage and Current Laws apply to all ac circuits?
Kirchoff's Voltage and Current Laws apply to all AC circuits as well as DC circuits. Other laws, such as Ohm's law and Norton and Thevanin equivalents apply equally as well. The complicating factor is that, at AC, current and voltage are not usually in phase with each other, unless it is a simple resistive circuit. That makes the math harder, but it does not make it invalid or impossible.
