When ever we use any power supply, we must take care of the ratings of the power supply. If the system requires more than what is required then we may face many problems:
CASE : When current extracted from the power source is greater than maximum current rating
Example1: Suppose we have a battery of voltage 5V and current ratings of 20mA. If we use this battery with a load of 200 ohms as shown then would it work properly?
If we calculate the amount of current in the circuit, then Icircuit = 5/200 = 25 mA but the current rating of the battery is 20 mA which is less than the current in the circuit.
Icircuit = 5/200 = 25 mA > (I max= 20mA)
Hence as the current being extracted from the battery is greater than the current rating of battery, the voltage provided by the battery drops. If we check the voltage provided by the battery in the circuit, it would be much less than 5V because battery can provide fixed amount of power
Power = V max * I max = Constant = Vcircuit * Icircuit
When the Icircuit is greater than I max then Vcircuit would become less than V max=5V
Example2: Suppose we have a battery of voltage 9V and current ratings of 20mA. If we use this battery to run a motor which requires the current of 30mA as shown then would it work properly?
As the current rating of the battery is less than the current requirement of the battery, hence if we check the voltage supplied by the battery then it would drop below 9V.
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Now I amm going away to ddo my breakfast, aterward havving mmy breakfast
coming again too rewd otgher news.
I don’t think the title of your article matches the content lol. Just kidding, mainly because I had some doubts after reading the article.
Thanks for the clear explanation! The example made it really easy to understand how exceeding the current rating of a power supply affects voltage. I never realized how critical this balance is for proper circuit performance. Great post — looking forward to reading more from the “Project Basics” section.
Great breakdown! The example made it super easy to grasp how overload impacts voltage. Learned something new—thanks for sharing!