Hi, i am working on solar energy. I have seen a solar energy cell generates about 10 micro ampere of current. We have this formula, V = Ir . Therefore I could produce a 9v worth of energy source. What i have to do is connect the energy source (10uA) to a 1 megaohm resistor. It will produce about 10V, which is what i want. Is this possible? If yes, i would like to implement it in such a way.
ya that teacher really needs to go back to school.
the best way u can go about this is by adding other solar panels and connecting them in parallel because when they are connected in parallel, the current will be added together while the voltage remain almost constant.
In theory that would be true for a perfect current source (one that can force a certain current through ANY resistance no matter how large. This is the same as saying the current source can produce an infinite voltage difference). In reality, there are limits. A solar cell, for example, might produce 10uA in short-circuit (at 0V) and produce 3V at open circuit (0A). Anything in between is less than 10uA and less than 3V.
I dont get what you guys are saying. Come on, the formula is v=ir. where v is voltage, i is current and r is resistance. So, R is directly proportional to v. If R is increased, then V is also increased.
I dont get what you guys are saying. Come on, the formula is v=ir. where v is voltage, i is current and r is resistance. So, R is directly proportional to v. If R is increased, then V is also increased.
I dont get what you guys are saying. Come on, the formula is v=ir. where v is voltage, i is current and r is resistance. So, R is directly proportional to v. If R is increased, then V is also increased.
You are thinking about it backwards. What you solved is how much voltage the solar panel would have to produce to push 10uA through a 1MΩ
resistor.
Only active components produce power, which is the solar cell itself. The resistor is a passive component which can only dissipate power, not enhance it. If R is increased, then V MUST increase from the ACTIVE source to push the same current.
What are you connecting that needs 10V? 10ma is not much current and that's probably only available in full sunlight at noon near the equator. What's the model of the panel you're using?
Sorry Blueroom but From what the OP was saying the "panel" is producing 10µA not 10mA
Nick If you are having difficulty think about it this way.
In your mind replace your panel with a 1.5v battery that can supply 10µA. OK
Now it does not matter what size resistor you connect in the circuit you can not have more than 1.5v
Basically what we are saying is that yes V=IR R=V/I I=V/R but in your case V is a constant. Change R and I will change but unless R is low or short circuit V remains constant. (Purists I know allow for light on the solar cell) but basically this is correct.
I dont have a solar panel yet. I read online on how to make one. So, it produces about 10uA. Transformer cannot change the voltage too, can it? I want to change the voltage because i want to use it with my circuit which requires 10V.
At ten millionths of an amp your not going to even have enough power to light up a high brightness LED even to a dim state.
Sorry but the ambient static electricity you generate when walking over carpet has more power available.
Hi, i am working on solar energy. I have seen a solar energy cell generates about 10 micro ampere of current. We have this formula, V = Ir . Therefore I could produce a 9v worth of energy source. What i have to do is connect the energy source (10uA) to a 1 megaohm resistor. It will produce about 10V, which is what i want. Is this possible? If yes, i would like to implement it in such a way.
Considering the fact you are mixing terms for electromotive force (volts) and current (Amps) and calling them energy, I think you need to go back and ask the teacher some questions about electricity.
Ok. I am studying more now. I have a question. Lets say i have a 3v and 0.5A solar panel. How can i increase the voltage. Won't a transformer work for this? I know voltage is essential in maximising the efficiency of a device (motor). I don't want to increase the current as it will negatively affect the efficiency of the device (motor).