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marciokoko
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Im reading this stack exchange question:
http://electronics.stackexchange.co...ochoosevalueofresistorinvoltagedivider
So there is a 5V circuit with 2100 Ohm resistors that yields a voltage drop across the divide of 2.5V. The current through the circuit is 25mA.
There is a second circuit the same 5V potential with 21M Ohm that yields a voltage drop across the divide of 2.5V. The current through the circuit is 2.5uA.
That is due to the fact we added those resistors. The larger the resistor, the smaller the current because more electrical energy is wasted as heat energy.
Then he goes on to explain that he adds a load to the first circuit, a 1K Ohm loadresistor to what used to be a 2.5V line. Now the potential is 2.38V instead of 2.5V and it has a 2.38mA current through it. It explains because: because the two bottom resistors are in parallel. I don't understand why. I know the total resistance in a circuit with parallel resistors is 1/Rt = 1/R1 + 1/R2 but how do I get 2.38mV and 2.38mA?
This is what I came up with:
http://electronics.stackexchange.co...ochoosevalueofresistorinvoltagedivider
So there is a 5V circuit with 2100 Ohm resistors that yields a voltage drop across the divide of 2.5V. The current through the circuit is 25mA.
There is a second circuit the same 5V potential with 21M Ohm that yields a voltage drop across the divide of 2.5V. The current through the circuit is 2.5uA.
That is due to the fact we added those resistors. The larger the resistor, the smaller the current because more electrical energy is wasted as heat energy.
Then he goes on to explain that he adds a load to the first circuit, a 1K Ohm loadresistor to what used to be a 2.5V line. Now the potential is 2.38V instead of 2.5V and it has a 2.38mA current through it. It explains because: because the two bottom resistors are in parallel. I don't understand why. I know the total resistance in a circuit with parallel resistors is 1/Rt = 1/R1 + 1/R2 but how do I get 2.38mV and 2.38mA?
This is what I came up with:
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