I'd like to understand this simple schematic to see if I get things clearly

alt text

If I understand things correctly:

  • when the switch is open, the current at the right black point is zero, the voltage is 12V.
  • when the switch is closed, the current at the right black point is 20 mA, the voltage is 0V

If this is correct, suppose now a similar schema with the following difference: between the switch and ground I have another 600 ohm resistor.

  • the potential at the black dot with the closed switch is now 6 V ?
  • any point along the horizontal connection between the T junction and the black dot are at the same potential. I guess they also enjoy the same current. If two points in a circuit experience the same potential and they are directly connected, do they always experience the same current as well ?

3 Answers 3


(In ideal conditions) You're correct in your understanding of the first case. When you close the switch, you short one end of the resistor to ground, so there's a 12V drop across it and I=0.02A.

In your second case, with the switch closed, you are creating a voltage divider so the point between the resistors is at 6V.

If you don't have a load connected along Vout (between the black dot and ground), then I=0 there since there isn't a voltage drop along the horizontal connection.

Measuring along the path connecting the resistors/switch, you'll see your 0.01A, and since there is no other path for the current to take, both resistors will see the full 0.01A.


I see where you are coming from, but you have 1 concept wrong.

From what you show, it is unknown what happens at vout. If you just have it connecting to something that is measuring the voltage, say a dmm or a microcontroller, then you can assume you have a very large resistor from vout to ground.

So case that the switch is open, you have 600ohms in series with a very large resistance. V/R=I, but since R is 600+verylarge, I is almost 0.

Now in the case the switch is closed, you have a wire (0ohms) in parallel with with very large, which results in an effective resistance of 0. So now you have V/R=I > 12/(600+0)=.02V

Now you can step back and look at voltages:

For the case that the switch is open, you have little current through the 600ohm resistor so V=0*600ohm=0v drop across the 600 ohm resistor, so Vout is 12v.

  • \$\begingroup\$ I work in an ideal condition where measure does not perturb the system \$\endgroup\$ Commented Nov 2, 2010 at 16:44
  • \$\begingroup\$ That is fine. In ideal conditions your DMM resistance will be infinity. In non-ideal it will be less then that. \$\endgroup\$
    – Kellenjb
    Commented Nov 2, 2010 at 18:01
  • \$\begingroup\$ In the case that your switch is open and you assume there is nothing connected to your output, then you don't have a circuit. A circuit by definition creates a complete loop. If you assume a resistance of infinity you can close your loop and use infinity in your standard circuits calculations. \$\endgroup\$
    – Kellenjb
    Commented Nov 2, 2010 at 18:19

"I guess they also enjoy the same current."

No, that's wrong. Your understanding of the current through the resistor with switch open and closed is correct. With nothing connected to the output the current only goes through resistor and switch. So the switch sees the same 20mA the resistor sees. When you place another 600\$\Omega\$ in series with the switch that will be 10mA. Let's keep that resistor there.

This circuit in itself is useless, you always want to connect something to the output. Let's assume that's a circuit which can represented by a 1200\$\Omega\$ resistor to ground. Then, with the switch open the output will be 8V, and there will flow 6.67mA into it. Zero through the switch. When you close the switch the 1200\$\Omega\$ becomes parallel to the switch's 600\$\Omega\$ giving an equivalent resistance of 400\$\Omega\$. The total current through the top resistance will then be 12V/(600\$\Omega\$ + 400\$\Omega\$) = 12mA, causing a 7.2V drop across the top resistor, so that the output level is 12V - 7.2V = 4.8V.

Now you're right that all components on that horizontal line to the output will see that same voltage, but they don't share the same current. Part of the 12mA will go through the switch with its resistor, part will go in the output's 1200\$\Omega\$. The distribution of the current is inversely proportional to the resistance of the different paths. The switch will draw 4.8V/600\$\Omega\$ = 8mA, the load will draw 4.8V/1200\$\Omega\$ = 4mA.


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