I am noobi in soldering and don't have much knowledge to identify components, i have a lot of Capacitors and resisters, now i want to know if i need 10U 25V Capacitor from what i have stock than how i check and select correct one, if Digital Multi Meter set to 200 Ohms range than what ohms reading it will show if i want 10U 25V? in case if i don't have 10U 25V than what will be alternate, i am trying to fix my laptop motherboard charging section so need to replace 10U 25V Capacitor, below is 10U 25V Capacitor which i need to replace from motherboard, any help will be apprenticed, Thanks.
closed as unclear what you're asking by Chris Stratton, pipe, Lior Bilia, Voltage Spike, Bimpelrekkie Nov 16 '17 at 6:49
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When you test a capacitor with an ohmeter you will get a measurement. But it's very hard to correlate the ohmeter reading with the capacitance of the capacitor, so the measurement is not very useful.
What you read with the ohmeter depends on the details of the design of your multimeter (how much current or voltage it uses for its resistance measurement) and will typically depend on how long since you contacted the capacitor with the probes, and the state of charge on the capacitor when you made contact.
If you measure the capacitor without removing it from the circuit board, it will also depend on all the other components connected to the same nodes on the board.
A capacitors' resistance is reactive. The resistance measurement will depend on a few factors. The formula for capacitive reactance is
Xc = 1/(2 pi F C). Where "Xc" is capacitive reactance (resistance), "pi" is the number pi, "F" is the frequency of the signal through the capacitor, and "C" is the measurement of the capacitor in Farads. In other words, you can't just put a multi meter on a capacitor, and measure it's ohm's. The Ohm reading of a capacitor is dependent upon the frequency going through it and it's capacitance value.
But, to answer your question, If the new capacitor that you have is marked "10U 25V" properly and the markings are correct, it should work. It should be somewhat safe to assume that the original designer used the proper component when the circuit was originally designed.