Voltage and resistance are distinctly different quantities. I can say some voltages have very low resistances (and high currents), such as when you drop a hair dryer into the tub. You're question as it is written does not make sense.
Resistance of a short circuit is actually very low, ideally it is 0 Ohms. In practice the resistance of a short circuit will be equal to whatever the resistance of the short circuited wires is, which is typically very small. So if you substitute very small resistance value R into Ohm's Law(I = V/R), you will get a very high current flowing. Where V=voltage,I=current.
Voltage = (current) x (resistance) Current = (voltage)/(resistance) Resistance = (voltage)/(current)
An ideal amplifier will have a very high input resistance, and low output resistance. This is so it doesn't "load" the input circuitry, and can drive output circuitry. Say you hooked an op amp up to a microphone that kicked out 100mV unloaded. If you shorted out the microphone, the output would approach 0mV (very low resistance load). If you connected a very high resistance load instead, the output would remain close to 100mV. As the load becomes a smaller resistance/bigger load, the output voltage will decrease until it is no longer useful. This is because the microphone has an internal resistance that remains constant while you're varying the attached load. The voltage drop across the internal resistance will increase as current output increases, causing less of a signal to be delivered to the amplifier.
the common emitter configuration is most widely used in amplifer circuits because of its high voltage,current & power gain.the common emitter configuration is most widely used in amplifer circuits because of its high voltage,current & power gain.
no it is not possibleAnswerYes, by changing the voltage OR the resistance.
ammeter connect in series in circuit to measure the current, if the Ammeter have a high resistance it would effect the voltage value because there will be a drop voltage over the internal resistance of the Ammeter, so we desgin the ammeters with very low resistance...
A megger measures very high resistance to check insulation, therefore it must use very high voltages so as to detect the very small current flowing.
The electric resistance is related to the diameter and extension of the wire submitted to a determined voltage which will determine the electric current flowing into the wire.AnswerVoltage has no effect on resistance. Resistance is determined by the length, cross-sectional area, and resistivity of a material (resistivity is affected by temperature, so temperature indirectly affect resistance).
It can connect 2 pins with a very high resistance and with a quite low resistance, depending on the voltage of the corresponding input pin.
Resistance of a short circuit is actually very low, ideally it is 0 Ohms. In practice the resistance of a short circuit will be equal to whatever the resistance of the short circuited wires is, which is typically very small. So if you substitute very small resistance value R into Ohm's Law(I = V/R), you will get a very high current flowing. Where V=voltage,I=current.
In electrical systems, voltage and current are related by Ohm's Law, which states that voltage equals current multiplied by resistance. Therefore, high voltage does not necessarily mean high current, as the current also depends on the resistance in the circuit.
By overhead cables, supported on masts, at very high voltage, to reduce the effects of resistance in the wires.
Use a voltage divider and an standard high input impedance voltmeter connected to the low voltage output tap of the divider. Just check that: a) the voltage divider has enough resistance to minimize loading of the voltage source (The voltage divider resistance should be as high as possible). b) the voltmeter's input impedance is at least 10-20 times larger than the output resistance (impedance) of the divider. If necessary add a high-input impedance amplifier or a transducer between the divider output and the voltmeter. What is high voltage for you?
high voltage - this minimizes I^2*R losses (losses due to the inherent resistance of the wire).
There are two ways of looking at this question, depending on what you mean by 'voltage'.The first applies to the supply voltage, which is quite independent of a circuit's load resistance. In other words, changing the load resistance will have no effect on the supply voltage (within limits; for example, and extremely-low resistance might cause the supply voltage to collapse!).The second applies to any voltage drops, which are proportional to the resistance across which they appear. If, for example, you have a high resistance and a low resistance, in series, then the higher voltage drop will appear across the higher resistance.
Voltage = (current) x (resistance) Current = (voltage)/(resistance) Resistance = (voltage)/(current)
Voltage = (current) x (resistance) Current = (voltage)/(resistance) Resistance = (voltage)/(current)