But a reader asked something different. My issue is with the delta It seems the answer to 1 is V since deltaV is V. Are we even using deltaV or just the electric potential for that point, in which case it is zero for 2? The delta in the equation referenced is part of the work-energy theorem -- work done by an external force is equal to the change in kinetic plus change in potential energy. Here, the potential energy experienced by a charge q is qV, where V is the electric potential at a position.
The question that I think you mean to answer is, "How much work is necessary to move the object from A to B with no change in kinetic energy? The value of the electric potential itself is irrelevant And to change potential energy, the object has to move from one position to another where the electric potential has changed. Thinking anti-clockwise on the diagram, the different graph lines might depict a decreasing resistance e. However, this is only true, giving a linear graph if the temperature doesn't change.
Comments about variables in this particular Ohm's Law experiment. Current is always determined by a combination of the p. The independent variable is what we change or control in the experiment - in this case you can consider it as the p.
One convention is to plot the independent variable on the x-axis, and the dependent variable on the y-axis. This means the resistance R, is the reciprocal of the gradient - a bit more awkward to calculate the resistance than from the V versus I graph plot, where the gradient is the resistance. The dependent variable is what we are testing or measuring in the experiment, this is the current I A , which depends on the variable resistor setting, which in turn controls the potential difference across the resistor.
Investigating the current - voltage characteristics of a metal filament lamp. When electric charge flows through a high resistance, like the thin metal filament of a lamp, it transfers some of the electrical energy to the thermal energy store of the filament. The electric charge does work against the resistance. Circuit 45 shows how you can investigate the current - potential difference characteristics of a filament bulb. The voltmeter is wired in parallel with the thermistor, the p. V is measured in volts V.
The variable resistor allows you to vary the p. The ammeter, wired in series, gives you the current I reading in amps A. The passage of current heats up the filament and the rise in temperature causes the resistance to increase. So a filament lamp is a non-ohmic conductor. This 'heating up effect' affects all resistors. As the current increases, more heat energy is released and the filament gets hotter and hotter, so further increase in temperature further increases the resistance.
This decreases the rate at which the current increases with increase in potential difference. Therefore the gradient of the I-V graph curve decreases and increasingly so with increase in temperature - graph 2.
Its a non-linear graph. If the gradient is changing, then the resistance is changing. The graph 2 is constructed on a crosswire axis. When the current A is NOT proportional to the p. You get the same I-V shaped graph for a thermistor. Theory - with reference to the metallic structure diagram. A metal crystal lattice consists of immobile ions and freely moving electrons between them. As the temperature increases, the metal ions vibrate more strongly into which the electrons collide and this inhibits the passage of electrons - reducing the flow of charge.
As the current increases, the vibrations increase causing more of the electrical energy to be converted to heat - increasing the temperature AND the resistance of the metallic filament, thereby lowering the current even further. So, an i ncrease in temperature increases the resistance a filament lamp or most other resistors and lowers the current flowing for a given p. If a resistor becomes too hot, almost no current will flow.
There is one important exception to this 'rule', see notes on the thermistor where the resistance actually falls with increase in temperature. The filament bulb is just one of many examples were energy is transferred usefully , BUT there is always heat energy lost to the thermal energy store of the device and the surroundings. An inert gas such as argon or nitrogen is added to reduce this evaporation - any evaporated tungsten atoms hit the unreactive and so non-oxidising Ar or N 2 molecules and hopefully condense back on the filament.
See Conservation of e nergy, energy transfers-conversions, efficiency - calculations. Investigating the current - voltage characteristics of a diode. The current through a diode flows in one direction only - see graph 3. The resistance in the opposite direction is very high - hence effectively a 'one way' system.
Circuit 43 shows how you can investigate the current - potential difference characteristics of a diode. A diode has a very high resistance in the reverse direction. There is also a threshold p. Therefore you get a top right portion of the graph 3 compared to graphs 1 an 2 above. This is because when you do the experiment using the circuit described above, on reversing the connections, you will find no current flows as you vary the p.
The graph 3 is constructed on a crosswire axis. Since the current only flows one way through a diode, it can be used to convert an ac current into a dc current. Practical work to help develop your skills and understanding may have included the following:. In your course, you might not need every formula - that's up to you to find out. V the potential difference p.
Potential difference is the work done in moving a unit of charge. It indicates how much energy is transferred per unit charge when a charge moves between two points in a circuit e. In practical terms, wattage is the power produced or used per second. For example, a watt light bulb uses 60 joules per second.
You abbreviate a watt as simply W, so a watt light bulb converts joules of electrical energy into light and heat every second. Energy and power are closely related but are not the same physical quantity.
Energy is the ability to cause change; power is the rate energy is moved, or used. Electricity is measured in units of power called Watts, named to honor James Watt, the inventor of the steam engine. A Watt is the unit of electrical power equal to one ampere under the pressure of one volt. Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics.
It only takes a minute to sign up. Connect and share knowledge within a single location that is structured and easy to search. It then appears that you are doing some dimensional analysis on both sides of the equality. The units for both sides of the equation are energy. You haven't provided the context for your equation, but it appears to be related to the work energy theorem which states that the net work done on an object equals its change in kinetic energy.
The left side appears to be the kinetic energy acquired by the charge between the two points due to the work done on it, assuming the charge initially is at rest. Sign up to join this community.
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