I want to know whether current & resistance can cause a voltage across the circuit, according to Ohm's Law $V=IR$. Is it possible to use a low current to produce a high voltage?! How can this be, as current is directly proportional to voltage?
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1$\begingroup$ Please try to improve your formulation first and take the time to add more details if you expect an answer. $\endgroup$– EllieAug 5, 2014 at 12:38
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$\begingroup$ Related: physics.stackexchange.com/q/51875/2451 $\endgroup$– Qmechanic ♦Aug 6, 2014 at 18:37
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3 Answers
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For a constant resistance Ohm's Law is $$V = IR.$$ Now, it happens that it's pretty easy to make a constant-resistance device (we call them "resistors") and that it's easier to make constant-voltage devices than constant-current devices. So most of the circuit problems we encounter have a constant-voltage device like "a wall socket" or "a battery" or "a regulated power supply" driving currents through some loads. The reason for this is probably that a device with a constant voltage across its elements can still has that same voltage when it's removed from the circuit: if I have a 1.5 V battery, its potential is 1.5 V whether it's running my flashlight or sitting in my closet. A true constant-current device, on the other hand, would quickly build up a big charge on its terminals if you didn't provide some return path.
However, there's nothing special about voltage that makes it more fundamental than current, and there are several types of devices which can be thought of as "current sources." The most common is a bipolar junction transistor, a three-terminal device, when connected to an appropriate power supply network. If a small current $I$ flows into a transistor's "base", a much larger current $hI$ (with $h$ typically 200–300) will flow from the transistor's "collector" to its "emitter." Depending on the resistances between the power supply, the transistor, and ground, the transistor current may drive the voltage at the collector/emitter terminals to just about any value. (In digital logic, the transistors are usually driven "to saturation," so that the voltages are either near the supply voltage or near ground — effectively the transistor acts like a switch. And in real analog circuits, one uses transistor networks, since the gain $h$ is not a stable, reliable parameter for a single transistor.)
An even better approximation for a constant-current device is a photodiode. In the photoelectric effect, one absorbed photon causes the emission of one electron; when the device is under more light, it produces more current. A photodiode with no current return path essentially charges itself up like a capacitor until the voltage at the junction is the same as the work function for the cathode; a photodiode whose terminals are connected by a resistor will have an output voltage $V=IR$, where $I$ is the photocurrent. If the photodiode is powered, as in a vacuum photodiode or a photomultiplier, the output voltage can be quite large — or more frequently, a tiny photocurrent can produce a measurable voltage signal.
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Ohm's law can be summarized as this: The drop in electric potential (voltage) across a section of a circuit is equal to the total current flowing through that section of the circuit multiplied by the total resistance of the section of the circuit.
By that law, if you have a current flowing and there is some net resistance, there will be a drop in voltage.
The equation from Ohm's Law is $$V=IR\quad\text{or}\quad(Voltage)=(Current)\times(Resistance)$$ Based on this, it is certainly possible to have a very low current and a very high voltage. Remember, current is only directly proportional to voltage if the resistance stays constant. If resistance changes, then current does not need to change proportional to the voltage drop. Also, for example, if you have a large resistance (say 10 trillion Ohms), you can have a potential of 1 billion volts and your current would still only be 0.1mA (a very low current)
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$\begingroup$ Yes you can but ill require more than a simple resistance and dependeds if its AC or DC. tip: search for solenoids. Btw all fluorecents lamps uses a reactor to generate high voltage/low current since gases only ionizes at high voltages $\endgroup$– jeanAug 5, 2014 at 14:12
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$\begingroup$ Just adding. For a pratical experiment OP ill need to keep the current constant and increase the resistance to observe the increase in voltage and that implies in some pratical problems. At the other hand he can observe that in any fluorescent lamp or at his car eletric starter system =) $\endgroup$– jeanAug 5, 2014 at 14:50
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Let's get this straight, current flow in a conductor is a direct result of a voltage applied across it (a potential difference or electric field pushing the electrons through the wire), and no conductor has zero resistance (apart from a superconductor).
The current is a result of the voltage, not the other way around.
You can have a large voltage across the conductor with low current if the resistance is high. According to Kirchoff's loop law, the electric potential energy provided to the electrons must be dissipated by the circuit. Assuming the voltage is held constant, if the resistance is high, the current must reduce to compensate and hold V=IR true. This is how you can have a high voltage with a low current.
If you need any further clarification, leave a comment, however, I sense that if you revisit a few concepts this should become clear to you.
