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# Potential divider

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This is a simple potential divider...The battery supply is 12V...The voltage accross the 200 ohm resistor is 8V...and the voltage across the 100 ohm resistor is 4V...shouldnt the resistor with the greater resistance have the lower Voltage...i dont undertand this?...please help ##### Share on other sites

Hi Gugh,
Perhaps you will understand the operation of a voltage divider if you use extreme values for the resistors. Change the lowest value resistor to a dead short. Then change the highest value resistor to an open circuit (extremely high resistance) Now when you connect your voltmeter to its output and ground you will measure the maximum voltage that is available, right?
How much voltage will you measure across the dead short that doesn't have any current flow?

You can calculate the voltages by using Ohm's Law:
1) The current in the series resistors is 12 divided by (100 plus 200) which equals 40mA.
2) The voltage across the 100 ohm resistor is 40mA times 100 which equals 4V.
3) The voltage across the 200 ohm resistor is 40mA times 200 which equals 8V.
4) As a check, the voltage across both resistors is 40mA times (100 plus 200) which equals 12V.

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The current through both resistors is the same since they are in series. I = U/R . 12/300 = 0.04A. This is the current in the resistors. The voltage over the 200 ohm resistor U = R*I. 200*0.04 = 8Volts. I am sure you can calculate the second voltage yourself. ;D

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Now the reason for the voltage dividing. If you imagine that a small tube is the high value resistor and a large tube is the low value resistor you can see that the there requires more energy and thus more voltage across the small tube than the large tube to get the same coulombs per second. The electrons have an easier time moving in the large tube, which is the low resistance, and require less voltage or energy.

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