andrew8485 said:
so is what i am about to say right. After the 2 resistors in series (which will equal 100 ohm) the voltage will be 0.00000 I don't think I got what you were trying to say. Sorry for being a pain
OK -- here's the circuit that I described:
**broken link removed**
When measuring voltages at any point in a circuit, the reference must be clearly understood if you wish to obtain meaningful readings. Considering the circuit above, the measurements that I mentioned in my previous reply would be those taken
across each resistor, i.e. with the VM probes at point D and E for R1, and at points F and G for R2. If, on the other hand, you are measuring
with respect to chassis (ground), the expected voltages would be different -- they would in fact be the source voltage less the accumulated circuit drop to the specific point at which the measurenment is taken.
Watch this -- placing the VM - probe at point A or H and the VM + probe at point C would give an anticipated reading of 12V -- you are in effect measuring the drop across both R1 and R2. Move the + probe to point F, and the anticipated reading is now 6.36V, as you are measuring the drop across R2 only. Move the + probe to point D and the - probe to point E, and the anticipated reading is now 5.64V -- the drop across R1. With the - probe at point H and the + probe at point G, the anticipated reading would be 0V, with any reading other than that indicating an abnormal drop between those points. Similarly, if you connect the VM between points A and H, you should expect a 0V reading, with any other result showing a problem with the circuit.
This measuring technique is useful for locating opens and shorts in a circuit. Suppose the wire were broken between points C and D. What should the voltage read, with respect to ground, at point C? At point D? If the wire were intact, the voltage would be source voltage (12V) at both points. If the wire were broken, Point C would show the source voltage, while point D would show zero volts. In essence, an open is an infinite resistance, with full source voltage being dropped across it. As to locating shorts, consider the drop across R2 if the resistor were somehow shorted. With a direct short, there would be no drop across the ressitor, meaning that (in the sample circuit given) the circuit current would now increase to .255A (12/47), as the entire source voltage would be dropped by R1. The two hints here at a circuit problem are the higher-than-expected current (0.255A vs. 0.120A) and the lack of any voltage drop at R2. If the value of either resistor were to change, its effect on circuit current, and thus voltage dropped across the resistor, will also change.
This technique can thus be used to locate abnormal circuit resistances due to poor soldering, loose connections, undersized conductors, and so on. Each instance of sub-par circuit condition will likely cause a voltage drop at the location of the problem. For example, if you were to measure (with respect to ground) a different voltage at point E than at point F, you would have an indication of a voltage drop along the wire or PCB trace between those two points. While wire losses are always present to a certain extent, a well designed circuit will have minimal (extremely low) wire losses.