IMPORTANT DOUBT

Thread Starter

whale

Joined Dec 21, 2008
110
I CONNECTED +VE OF DC SOURCE(15V) TO A RESISTOR(100KΩ),WHEN I MEASURED POTENTIAL ACROSS -VE OF SOURCE AND ANOTHER END OF RESISTOR I COULD MEASURE ABOUT 13V,EVEN WHEN I MADE MY BODY AS RESISTOR I COULD GET 12V.
HOW CAN THIS HAPPEN?
WHEN WE CONSIDER -VE TERMINAL AT GROUND POTENTIAL,THEN +VE TERMINAL IS +15V(IT ACCEPTS ELECTRON AT A POTENTIAL 15V).

BUT WHEN THIS +VE TERMINAL IS CONNECTED TO HIGH RESISTANCE ,THE POTENTIAL BETWEEN RESISTOR ANOTHER END AND -VE OF SOURCE SHOULD REDUCE A LOT .BECAUSE IT ALLOWS ONLY FEW ELECTRONS TO PASS THROUGH.THUS THE DIFFERENCE IN ELECTRON LEVEL DECREASES THUS POTENTIAL SHOULD REDUCE.

WHY IT DONT HAPPEN?
 

SgtWookie

Joined Jul 17, 2007
22,230
It appears that your 15v source has a rather high internal impedance.
When you connect an external resistance across it's +V and -V terminals, you wind up with a voltage divider; some of the voltage is dropped across the internal resistance, and the rest is dropped across the external resistance.
 

Thread Starter

whale

Joined Dec 21, 2008
110
the source give 0.5A at short circuit.my question is how the electrons are accumulated the terminal of 10k ohm resistor,whose another terminal is connected to -ve of source.
with respect to +ve of source.
i can clearly identify that no current flow takes place. but how a 15v(between +ve and -ve)potential diff remains with less distrotation about 12v(between +ve and terminal of resistor,whose another terminal is connected to -ve of source)
 

SgtWookie

Joined Jul 17, 2007
22,230
You are not making sense. :confused:

If you connect a resistance across a voltage source, there will be a flow of current.
I = E/R, or Current(Amperes) = Voltage/Resistance(Ohms).
Therefore, if you place a 10k Ohm resistor across a 15v source (an ideal voltage source that has an impedance of zero Ohms), you will have 1.5mA (0.0015 Amperes) current flow through the resistor.
 
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