0
\$\begingroup\$

I've recently started reading through Practical Electronics for Inventors (Scherz and Monk 3rd edition), and have a few questions about a specific figure used when talking about grounding a circuit. Please bear with me.

enter image description here

The book notes how in 2.31a, there will be no current flowing from the power supply, as there is no current return path with the load being hooked up to the + and ground terminals of the power supply. In this case, wouldn't there still be some current flowing to ground until the supply runs out of juice, being that there's no return path for the electrons? There's still a very small amount of resistance in the wire correct? (I'm assuming in all these diagrams, the wiring for the ground and - leads are not actually connected, although the wires overlap in the diagram?)

Second, knowing that there has to be a return path for the current back to the supply, how is the current returning for the floating load? Looking at the left diagram, when the current goes through the load and has to go through the - terminal wire, does it go back all the way to the neutral wire, or does it loop through the transformer? What are the actual electrons doing in this situation, as opposed to conventional current flow?

Lastly, how is the jumper wire in the grounded load work exactly? Wouldn't the current seek straight to ground now that the jumper wire is there, like in 2.31a, rather than attempting to move across whatever internal resistance is within the rectifier/transformer?

I appreciate any help. Just trying to visualize these sorts of concepts.

\$\endgroup\$
1
\$\begingroup\$

In this case, wouldn't there still be some current flowing to ground until the supply runs out of juice, being that there's no return path for the electrons?

No, because as you have noted, there is no return path.

Second, knowing that there has to be a return path for the current back to the supply, how is the current returning for the floating load?

Because, as you note, there is a return path. A power supply will use a transformer for reasons of isolation and this, effectively, makes a floating voltage source.

Wouldn't the current seek straight to ground now that the jumper wire is there

Ground is not a viable return path for current - current will want to return to the supply source negative terminal.

Imagine that instead of planetary ground or earth, these circuits were on a space craft and the ground connection was just the space craft chassis - would you see things differently?

\$\endgroup\$
  • \$\begingroup\$ Thank you for responding Andy. Can you dive further into why the lack of a return path makes no current flow? I can guess that with no return path, that's there's no electron chain getting pushed back to the negative terminal of the power supply, but shouldn't there be some charge that's put through the wire in this case to ground, like a short circuit? \$\endgroup\$ – XZE1527 Jan 7 '17 at 19:28
  • \$\begingroup\$ Also for the second question, you mentioned the transformer is isolated from the load? I know that the transformer moves the voltage value down, and the rectifier converts to DC. In that picture I posted to the left, for the floating load scenario, what physical component of this circuit is allowing the current to loop back to the + side of the load, if the transformer/rectifier is isolated? \$\endgroup\$ – XZE1527 Jan 7 '17 at 19:40
  • \$\begingroup\$ No return path means no impetus for current to want to flow. Sure you can create a big voltage and spray electrons seemingly into the air but that is the power of like charges repelling. We don't want that to happen in most circuits so we keep voltages low and use insulation but, with no path for current to flow back to the negative terminal there is no continuous stream. \$\endgroup\$ – Andy aka Jan 7 '17 at 20:18
  • \$\begingroup\$ A transformer isolates secondary from primary and not from the load. I think you misquoted me! \$\endgroup\$ – Andy aka Jan 7 '17 at 20:19
0
\$\begingroup\$

In this case, wouldn't there still be some current flowing to ground until the supply runs out of juice, being that there's no return path for the electrons?

Have you learned about capacitance yet? If so, think of this as charging a very small capacitor: It doesn't take very much charge, e.g. a small current for a small time, to charge it up to the small number of volts from the supply. Then they're at equal voltages and the current stops.

If you haven't learned about capacitance, consider this like static electricity: When you push charge into something from which it can't leave, i.e. into a resistor without a connection at the other end, you're charging it up. Just a tiny bit of static electricity will charge a large object (i.e to) to thousands of volts. It takes a very tiny amount of charge to charge a small body (the resistor) to a few volts to match the power supply.

Rough calculation: charging 1pF (a rough estimate of the resistor) to 10V takes 1 milliamp for a 10 nanoseconds. You won't see that.

\$\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.