I was checking falstad and came around this resistor circuit which lost me completely :

enter image description here

Knowing that input voltage is 5v and current is 17mA :

1- Why is the voltage 0 after the 200Ohm resistor knowing that It's 3.4V before It ?

2- Why is the current flowing without voltage ? (here is the animation for those interested

3- When the current arrives at the first junction, 15mA take the first route, and 2mA take the 2nd route, why is that ?

4- Check this 2nd circuit :

enter image description here

Why closing the right bottom switch drops the voltage to 0 after the 100/400/800 resistors ?

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    \$\begingroup\$ 1. Because it's tied to ground. 2. Why not? Does not violate any laws. 3. To minimize the energy loss. 4. Again, tied to ground. \$\endgroup\$ – winny Oct 1 '19 at 13:29
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    \$\begingroup\$ the voltage-meter needs a reference point; we often call that "GROUND" and define that voltage to be zero. \$\endgroup\$ – analogsystemsrf Oct 1 '19 at 14:25
  1. The voltage is zero after the resistor because you have defined it to be zero -- it is the circuit common (or ground).

  2. If you are talking about the RLC circuit on the front page, it is because the L and C are energy storage elements that are charged by the voltage source before the start of the simulation. The simulation begins at the opening of the switch.

  3. It is a parallel circuit. The current is inversely proportional to the resistance/impedance of each branch.

  4. The lower right switch is shorting out the 200 ohm resistor, making the voltage difference across it zero.

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