What's the Difference Between Potential Difference and EMF in a Solar Cell?

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The discussion revolves around a student struggling with concepts in engineering physics related to a solar cell's potential difference, internal resistance, and efficiency. The key question is the distinction between potential difference and electromotive force (emf), with clarification that emf is the voltage when no current flows, while potential difference accounts for internal resistance. The student seeks guidance on calculating the internal resistance and emf based on given voltages across different resistors, as well as how to determine efficiency. Efficiency is defined as the ratio of useful energy output to energy input, requiring calculations of both solar energy input and energy delivered to the circuit. Overall, the student expresses frustration with the physics course and a lack of understanding of the material.
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ok, I am taking engineering physics II ... unfortunately, I am forced to take this as part of my COMPUTER SCIENCE degree ... (why i have no idea..stupid, as i don't really care how many Newtons of force it takes me to push a key on the keyboard, and how much current it takes to go from the keyboard to the computer to put the key on the screen) ... sorry, just I am not the most science-oriented person. which is why I am not a computer engineer.

so i have a question.

the question on my homework is this:

A sloar cell generates a potential difference of 0.10V when a 500-Ohm resistor is connected across it and a potential difference of 0.16V when a 1000-Ohm resistor is substituted. What are (a) the internal resistance and (b) the emf of the sloar cell? (c) The area of the cell is 5.0 (cm^2). What is the efficiency of the cell for converting light energy to internal energy in the 1000-Ohm external resistor?

question. What is the difference between the potential difference, and its emf? i know V=IR . isn't the total V what comes out of the power source? and isn't the emf the power source? I'm lost on what this is.

also, for part c ... how would i approach in solving this? The chapter deals with the circuits, resistors, etc. but effeciency?

any help would be appreciated!
thanks
 
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Originally posted by JaeSun


A sloar cell generates a potential difference of 0.10V when a 500-Ohm resistor is connected across it and a potential difference of 0.16V when a 1000-Ohm resistor is substituted. What are (a) the internal resistance and (b) the emf of the sloar cell? (c) The area of the cell is 5.0 (cm^2). What is the efficiency of the cell for converting light energy to internal energy in the 1000-Ohm external resistor?

question. What is the difference between the potential difference, and its emf? i know V=IR . isn't the total V what comes out of the power source? and isn't the emf the power source? I'm lost on what this is.

also, for part c ... how would i approach in solving this? The chapter deals with the circuits, resistors, etc. but effeciency?

thanks

The EMF can be thought of as the Potential Difference at the terminals of the cell WHEN NO CURRENT is flowing. However, the cell has some resistance, so if a current flows through it, then there is a voltage drop inside the cell. The voltage at the terminals will then drop. The emf is the value without this voltage drop, the PD is the Voltage at the terminals when the voltage has dropped (start you car with the lights on and see them dim as the starter turns - that is internal resistance at work). Draw a diagram of the system, including a resistor next to the cell (the internal resistance, r) and then use simple circuit rules to solve it.

Efficiency is the ratio of useful energy out, divided by energy into a system. Work out the solar energy total into the cell and then the energy out.

Hope this helps.
 


Originally posted by Adrian Baker


Efficiency is the ratio of useful energy out, divided by energy into a system. Work out the solar energy total into the cell and then the energy out.

Hope this helps.

i am somewhat lost on the last part ... i am going to think that the energy in is 10 mW/(cm^2) ...

but i am lost on how to get the energy going out into the circuit ...

ugh, i hate my physics class...im so lost everyday ...
 
I think it's easist first to watch a short vidio clip I find these videos very relaxing to watch .. I got to thinking is this being done in the most efficient way? The sand has to be suspended in the water to move it to the outlet ... The faster the water , the more turbulance and the sand stays suspended, so it seems to me the rule of thumb is the hose be aimed towards the outlet at all times .. Many times the workers hit the sand directly which will greatly reduce the water...
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