One. Measure the load resistance. Divide by five. Use that as the added series resistance. (For 1000 ohm load, add 200 ohms in series.)
Note that this answer may be wrong for situations where the load resistance changes or the current is high, as you must consider source to load rejection ratio and power consumed by the added resistor.
Another way is to to get a stabilised plug top power unit . Open it up and use the electronic regulator from it. Wire your 6 volts DC in place of the transformer secondary and use output as normal. The integrated three terminal regulator you are looking for is called a 7805 industry standard 1 amp regulator.
ANSWER: Must know the load current first then add a series resistor to drop 1 volts with the current flow.
a; find the load current first them add a resistor in series to drop 1 volt
depends on the voltage of the batteries.. four 12 volt car batteries would output 4x12 = 48 v
You must hook them up in series. You will need four 12 volt batteries to do this.
Yes. The equivalent resistance of resistors in parallel is written as 1/Req=1/R1+1/R2+1/R3+... which, in this case, would be 1/Req=1/1000+1/1000+1/1000+1/1000=0.004. This means that Req=1/0.004=250Ohms.
depending on your load resistance what are you using . suppose you are using 100 ohm load resistance than current will flow 1.5*4=6/100=0.06Amp and if you are looking for battery current capacity than mind it you are using series connection the current will not change remain same as one battery but voltage will change as i mention above 6 volt.
The GCF is 4. The equivalent fraction is 1/2.
7 ohms (5+1+1)
A Kelvin bridge (as well called a Kelvin double bridge and in some countries a Thomson bridge) is a measuring instrument used to measure unknown electrical resistors below 1 ohm. It is specifically designed to measure resistors that are built as four terminal resistors.
depends on the voltage of the batteries.. four 12 volt car batteries would output 4x12 = 48 v
Four Dead Batteries - 2004 is rated/received certificates of: USA:R
To maintain the 12 volts using four batteries they have to be wired in parallel connections. This means that all of the positive posts are connected together and all of the negative posts are connected together. The total sum of all of the batteries will equal 12 volts.
2000 ohms
By deleting the first and last letter of FIVE it becomes IV which is a Roman numeral for four.
Two eight-ohm resistors in series would have a total resistance of 16 ohms. Two eight-ohm resistors in parallel would have a total resistance of four ohms.
C batteries use 1.5 volts. The number of amps depends on what device it is hooked up to. An average for four C batteries would be about 16 amps.
yes, it is normal. but you can reduce it by using ice packs
It depends on the values of the individual resistors. But if each resistor is identical, then the total resistance will be one-quarter that of an individual resistor.
What would the measured ohms be for two 100 ohm resistors wired in series? Two 100 ohm resistors wired in series measure 200 ohms.