Question about how resistors work

[Nerdydude101]

So i have recently been learning about electronics and i have a question about the way a resistor resists electric current. So i have a simple circuit, 9v goes into a 10k resistor, and then into a dc milliamp meter (its the one on the electronics learning lab from radioshack) and then to another 10k resistor and then ground. My question is how does the current drop from a resistor that resists current after the current has gone through the meter? shouldn't it resist 10k before the meter and another 10k after the meter? Thanks for the help!

 

[Gryd3]

So i have recently been learning about electronics and i have a question about the way a resistor resists electric current. So i have a simple circuit, 9v goes into a 10k resistor, and then into a dc milliamp meter (its the one on the electronics learning lab from radioshack) and then to another 10k resistor and then ground. My question is how does the current drop from a resistor that resists current after the current has gone through the meter? shouldn't it resist 10k before the meter and another 10k after the meter? Thanks for the help!

Your thinking of electricity travelling through the wire like a slinky... getting backed up only where there is resistance. Try to think of it like a more solid object.
Classroom hallways get pretty packed sometimes... if everyone needs to pass through a single set of doors, everyone slows down, not just the people in the doorway.

 

[shumifan50]

The relationship between resistance, potential and current is defined by Ohm's law I= V/R or V=IR or R=V/I. Inserting a resistor in a circuit causes a voltage drop across the resistor. Te voltage drop happens over each resistive component in the circuit and will be different based on the resistance(Ohm's law). However, the current through a single trace of the circuit is the same everywhere.
In your case, ignoring the resistance of the milli-amp meter which is negligible compared to the 2 x 10K resistors, you can calculate the current to expect as I=V/R =9/20000 = 0.00045A. That will be the current at all points in the circuit.
You can now calculate the voltage drop over each resistor: V=IR = 0.00045 * 10000 = 4.5V. This is sort of expected as the 9V will be an equal voltage drop over each resistor.

 

[BobK]

The three elements (two resistors and meter) are connected in series. Components in series always have the same current flowing through them.

Bob