Resistors are used to drop current Amp and Voltage, following Ohm's law.

I can't understand some of their applications: we use them cause we sometimes want to drop the voltage a little in a specific point of the circuit, but from what I understood the intensity drop they produce is system-wide.

So, in a circuit like this,


simulate this circuit – Schematic created using CircuitLab

the V5 emf provides X Amp of current to the diode (D1) and R1 provides a certain Voltage drop.

If I rearrange the diode to look like this, does the same current X flow through the diode D1? (therefore it is not able to draw all the current (charge) from the source?)


simulate this circuit

  • \$\begingroup\$ These circuits are functioning in an exact same way. And you might want to read a basic electronic circuits book (no offence intended). \$\endgroup\$ – Eugene Sh. Oct 22 '15 at 16:56
  • \$\begingroup\$ yeh that's why I'm here, I just don't get if resistors work that way or another \$\endgroup\$ – Massimo Pesavento Oct 22 '15 at 17:02
  • 1
    \$\begingroup\$ Generally: In a series-connected components the current is the same. In parallel-connected components the voltage is the same. That's it. The order they are connected doesn't matter. \$\endgroup\$ – Eugene Sh. Oct 22 '15 at 17:11
  • \$\begingroup\$ You might want to start with a simpler circuit. Replace the diode with a 1Ω resistor, so there is 100Ω in series with 1Ω. Only the battery can create or consume current, so the current through the 100Ω and 1Ω must be the same. The voltage drop across the two resistors is 5V, the current through each is the same. For Ohms law to be true, the voltage across the 100Ω resistor must be 100x bigger than the voltage across the 1Ω, but it doesn't matter which is first. The same idea applies to the diode and resistor. \$\endgroup\$ – gbulmer Oct 22 '15 at 17:48
  • \$\begingroup\$ Get yourself a breadboard and build the circuit. You could also simulate the circuit in a software like LTSpice. You can spend hours hypothesising about what might be going on in the circuit or simply try it out. With a simulator it's as complicated as doing what you did to create the graphics in your question and hit the "simulate" button. Learning humans should be operated in a closed loop configuration. \$\endgroup\$ – Magic Smoke Oct 22 '15 at 18:08

We sometimes use a water pump circuit analogy to help explain circuits. It's not a perfect analogy so don't take it too far ...

Think of the battery as a water pump. Think of the diode as a non-return valve (current can only flow in the direction of the arrow). Think of the resistor as a narrow piece of pipe which restricts or resists current flow.

The way you've inserted the diode it is forward biased so current will flow. How much current depends on the resistance. If we had no resistance (big fat pipe) we would get huge flow and the pump wouldn't be able to keep the pressure up (the voltage would collapse).

It should be fairly obvious in the analogy that it won't matter whether you put the non-return valve in the supply or in the return - the current will be the same.

So, provided we don't overload the battery (resistance too low) the 5V will remain fairly constant and the current controlled by the resistance. If you were able to measure the pressure / voltage at various points along the narrow pipe / resistor you would find that it is falling from 5 at the top to 0 at the bottom.

I hope that makes sense.

One other point that may help later: if the non-return valve has a spring in it to close it when there is no flow then it will take a certain amount of pressure to crack the valve open before there is any flow. As the pressure rises the valve will open more and more until fully open. Diodes behave in a similar fashion in that they usually require about 0.5 to 0.7 V before they will conduct. In your circuit this could be measured as 0.7 V across the diode and 4.3 V across the resistor.

To everyone else, I know the analogy isn't perfect so ...

  • \$\begingroup\$ Thank you for your answer, but I was not looking for an explanation of circuits, rather the fact that I didn't realize that my components were in series \$\endgroup\$ – Massimo Pesavento Oct 23 '15 at 11:37

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