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Sorry. Did not get you.
As you said, adding another load will change the effective resistance and hence the current being drawn from the source.
Hence now I have 20W of load.
What do you mean by the last line..?
"Load rating will not change. Instead the load will share the total power so...
I have a load of 10W, 75 Ohms connected to a DC supply. Now if I add another load of the same rating in parallel to this load, keeping the supply same. What will be the new load across the supply.
What if I add it in series instead of parallel..?
How does load rating change when connected in...
Found the problem, I was using a potential divider circuit to reduce the panel voltage for the ADC of micro-controller. It was the voltage of the potential divider that I had been looking at all the while. Upon scaling it with the same factor that the circuit had reduced it the voltage seems...
Ya there are a lot of fluctuations. Current value goes to a max of .57 which is fine but the problem is voltage, it goes to a max of 2.59V.
What bothers me is that power values that I get, it goes to a max of 1.45W. I expected the panel to give a higher output. Its it okay for a 10 W panel to...
I am using a 10W solar panel for a sun tracker project. I need to monitor the power at regular intervals.
But the voltage across the panel drops when I connect the load. The drop is in the range of 15V to 4V/5V.
The load that i am connecting is 2 10W resistors of 75ohms in parallel i.e 37.5...
Laptops have Li-Ion batteries. These batteries come with internal protection circuitry. So, can a particular usage pattern affect it health and ultimately its life?
As far as I get this, the source creates a potential difference in the path so that electrons can flow, just a water flows from a height to the ground. And, the electrons have potential energy in them, which they loose when they come across a load, of any kind.
Am I right? Does this explanation...
Let's say we have a source of electrical energy, in form of a battery or some other source. We then connect a load to it. The source creates a potential difference across the load and drives a certain amount of current through it. The amount of current depends on the load. We know that, current...
Let's say we have a light bulb of the following rating : 20W, 100V, 20 Ohms.
Let's say that, 20W of power is being supplied to the bulb.
By Ohm's Law current through the bulb will be 100/20 A.
Now if I use the power formula P=VI, I=20/100 A.
Again if I use the heat loss formula P=I^2*R, I will...
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