0
\$\begingroup\$

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 ?

\$\endgroup\$
  • 1
    \$\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 at 13:29
  • 1
    \$\begingroup\$ the voltage-meter needs a reference point; we often call that "GROUND" and define that voltage to be zero. \$\endgroup\$ – analogsystemsrf Oct 1 at 14:25
5
\$\begingroup\$
  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.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.