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About LinkBacksI just want to have a simple single circuit that uses various amounts of the battery's voltage. How can I make sure that the rest of the power left over does not short the battery while the circuit is closed?
September 20th, 2011, 06:10 PM
I just want to have a simple single circuit that uses various amounts of the battery's voltage. How can I make sure that the rest of the power left over does not short the battery while the circuit is closed?
I know Ohm's law, but how do I resist the left over power so there is minimal battery shortage. For example, if I have a series circuit and a 9 volt battery source with 0.2Ah, and I am consuming randomly from 0.00V to 9.00V, how can I make sure that the resistor will only consume what is not being used, so I don't short the battery?Originally Posted by MeteorWayne
There's really no way to answer your question, since it makes no sense. Sorry.
You may know Ohms law, but it is clear you do not understand it. Sorry.
Ohms law R = r L/A
That is not Ohm's law. That equation is to find the resistance of a wire given physical properties and dimensions, which has nothing to do with the original question.
What you need to find out is the impedance of your circuit. You might be able to make use of Thevenin's theorem to get this.
Then...
9v/ equivalent R = current. (Ohm's law)
0.2Ah / current = roughly how long your battery will last.
What kind of device do I need so that the battery does not short when the electromagnet accidentally drops the weight it's holding?That is not Ohm's law. That equation is to find the resistance of a wire given physical properties and dimensions, which has nothing to do with the original question.
What you need to find out is the impedance of your circuit. You might be able to make use of Thevenin's theorem to get this.
Then...
9v/ equivalent R = current. (Ohm's law)
0.2Ah / current = roughly how long your battery will last.
I don't get it. An electromagnet will draw identical current whether or not it's holding a weight. Maybe you could give us a more detailed description of your circuit and then tell us what you think is happening.
IMO use transistor. You can throttle the current flowing thru that transistor, and it won't short out if you want to use less current. The current flowing in the transistor is controlled by another (small) circuit that activate or de-activate the transistor.
If you use a resistor of variable ohm: the battery is still shorted out no matter how much voltage you use. Yes this is true.
No. It is not true.
To begin with, a circuit will always use all of a battery's available voltage.
A coil of wire will drain the battery with or without attracting/securing an object. So you want a means — apparently an automatic means — by which the circuit will disconnect from the battery when the coil does not have an object captured.
One solution. Build the coil around an open/air center. In its center place a steel core secured to the frame (but not the coil). Use a spring to push the coil forward away from an "NO" (normally open) switch placed in series with the coil.
Without an object in its grasp, the coil will remain forward, the NO switch will deactivate the coil (ie, no current flow), and the battery will not drain.
Pushing an object against the coil will activate the switch, thus closing the circuit and activating the coil to hold the object against the steel core (and keeping the coil pushed back).
Should the object be pulled away from the coil/core, the spring will push the coil forward, thus opening the switch and deactivating the coil.
Let me just get something straight. When the electromagnet is holding weight, is that more resistance than if it is not holding the weight? If yes, then aren't I shorting the battery out when the load is too heavy for the magnet and the load falls when the power is still on?To begin with, a circuit will always use all of a battery's available voltage.
A coil of wire will drain the battery with or without attracting/securing an object. So you want a means — apparently an automatic means — by which the circuit will disconnect from the battery when the coil does not have an object captured.
One solution. Build the coil around an open/air center. In its center place a steel core secured to the frame (but not the coil). Use a spring to push the coil forward away from an "NO" (normally open) switch placed in series with the coil.
Without an object in its grasp, the coil will remain forward, the NO switch will deactivate the coil (ie, no current flow), and the battery will not drain.
Pushing an object against the coil will activate the switch, thus closing the circuit and activating the coil to hold the object against the steel core (and keeping the coil pushed back).
Should the object be pulled away from the coil/core, the spring will push the coil forward, thus opening the switch and deactivating the coil.
Let's start by using the correct terminology. There is no short circuit. A short circuit is a condition where there is a fault in the cable between the power source and the load, in such a way that the current follows a "short" path from one terminal of the power supply to the other without going through the load. The "short" path may in some cases be physically longer, but is lower in resistance.
To answer the question, no the current will not be greater if the load falls off the electromagnet. An electromagnet is effectively an inductor. The inductance of a solenoid will change depending on the permeability of the core, and the permeability will change if the load falls off. This has no effect on the steady state current in a d-c coil.
The voltage across an inductor is given by L di/dt. The current when a voltage is applied to a circuit containing a resistance and inductance is an exponential function. When the voltage is first applied there is a high impedance due to the high rate of change in current. Over the long term as the current transient dies down (di/dt goes to zero) the voltage on the inductor is zero. The steady state current is limited only by the resistance of the coil.
Current will flow thru those resistor: thus represent losses, but if you use a transistor: you'd have a battery supplying variable current but without additional resistor.No. It is not true.
Losses in a resistor does not mean the same thing as "shorted out."Current will flow thru those resistor: thus represent losses, but if you use a transistor: you'd have a battery supplying variable current but without additional resistor.No. It is not true.
You may want/need to place a resistor in series with the coil so that, when the circuit is activated, the current will flow through the coil and resistor, thereby limiting the current drawn from the battery when the circuit is in a steady-state condition.
Otherwise, once the coil builds its magnetic field, its inductance drops to zero and the only impedance in the coil is the resistance of the wires, which can be very low, and this low resistance can draw high currents from the battery. This condition is what you are calling a short circuit.
Adding the resistor to the design limits the current to protect the battery from expending all its energy too quickly and/or permanently damaging it. Such a resistor is sometimes called a "current-limiting resistor" because its purpose is to limit the current to prevent for the reasons described above.
The B field is going to be proportional to the current going through the wire though. This is something to consider because if you increase the resistance by say a factor of 100, your magnet will be 100 times as weak. This can be fixed by having 100 times as many winds, which can get out of hand in a hurry.
I think that an analogy would help you. Think of voltage as not being stuff, but just push. Electrons are the stuff that gets pushed or pulled in a circuit. When electrons move either way in a circuit, that motion of those particles amounts to current. Power is the rate of using up energy. Power in a given circuit increases by the square of the current. So, voltage in a circuit is like water pressure in a pipe. Electric current around a circuit is like water flowing in a closed loop of plumbing.
Potential energy sort of remains within the battery except for what drain your circuit draws from it. When your circuit is closed, the resistance of the circuit almost completely determines how much energy is drawn from your battery, and thus determines the power being dissipated. The Ohms of resistance in a circuit make it sound as though the device is there to louse up your current, but it is there as the only path for any current. Everything else surrounding the battery terminals is (ideally) non conductive. The reciprocal of the Ohms value of a resistance is the conductance value (in mhos) that provides for passage of electrons. Without them, the battery is just resting on a shelf awaiting usage.
Someone may have warned you against getting a short circuit without describing what that is. Mostly, a short circuit would be what you call a big fat wire connecting the positive terminal of a battery to the negative terminal of the same battery. That ruins the battery in a hurry, and in many cases will cause your fingers to smell like roast pork and hurt. It can happen by accident especially if you wear any finger rings.
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