1
\$\begingroup\$

I am very new to electrical engineering, and one thing that has been bugging me is short circuits.

Main question:

When I have a circuit, how much current and voltage should be flowing back to my battery, safely (as not to short circuit, melt the wire, or damage components, or throw off voltage source equilibrium) with a load or resistance placed within the circuit?

I have recently spoken with a graduate electrical engineer (he has been out of the field for a while though) and he told me that if I have a circuit that allows voltage and current to flow back to the battery after resistance, then the battery will accumulate more voltage and current, thus sending more voltage and current back out to the circuit... This is also why a wire connecting both terminals of a battery with no resistance or any components that deplete the voltage and current will cause a short circuit. So am I supposed to design the circuit so that no voltage or current will be delivered back to the battery?

Example:

I guess a good example would be a circuit with a 5 Volt battery that has a 100 Ohm resistor and .05Amp (50 mA) current. Would this deliver voltage and current back to the battery after it passes through the resistor?

So then, in regards to the same example, if I have a 5 Volt Battery and .05 Amps flowing through the circuit, but a 50 Ohm resistor - would this short circuit?

Please let me know, as this has been bugging me a bit, and I think the issue is that i am thinking about it incorrectly...

I appreciate the help!

|improve this question
\$\endgroup\$
  • 7
    \$\begingroup\$ "... if I have a circuit that allows voltage and current to flow back to the battery after resistance, then the battery will accumulate more voltage and current, thus sending more voltage and current back out to the circuit ..." ... This is not someone you should be asking layman questions of. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 17 '18 at 19:07
  • \$\begingroup\$ Current would have to change direction to charge the battery \$\endgroup\$ – Sunnyskyguy EE75 Jan 17 '18 at 19:07
  • \$\begingroup\$ Haha, yea, definitely questioning the source, but I may have also butchered the hell out of what he was telling me. \$\endgroup\$ – seasidemango Jan 17 '18 at 19:36
4
\$\begingroup\$

Nice work on providing an example and a thorough question. You're obviously quite confused, but by talking it out in detail it's easy to see where you've gone wrong.

First thing, your colleague is not helping at all. Best to expel that advice.

Next, study this maxim:

  1. current flows through a circuit - the current in equals the current out
  2. voltage changes across components - the voltage goes from high to low around a circuit

So when you look at you battery connected with a resistor, 50mA flows out of the battery, 50mA flows through the resistor and 50mA flows back into the other terminal of the battery. Voltage starts high (5V) at the positive terminal, drops across the resistor and is low (0V) by the time it reaches the negative terminal. Physics makes it so.

Energy on the other hand, leaves the battery and is consumed by the resistor. The battery will not charge in this scenario, because the current direction is out of the positive terminal. The fact that it goes back into the negative terminal is a given (due to #1 above) and doesn't change anything.

A battery is a voltage source, so the current in the circuit is dictated by the resistance in that circuit. So if you have a 5V battery connected to a 50 Ohm load, there will be 100mA flowing through it - there's no argument there, no short circuit. Physics makes it so.

A short circuit only occurs when you put zero resistance across a voltage source.

Now you wonder, in a short circuit, how can the voltage go from 5V to 0V if there's no resistance in the circuit? The answer is that there's always something, even in a "short circuit". Say 0.01 Ohm. So you drop 5V across 0.01 Ohm and produce an enormous current, sparks, fire and all the fun stuff, and we call it a short circuit.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ This is excellent! Very well said sir. My assumption, then, is that a short circuit is also dictated by how much the component can withstand in regards to how much current flows through prior to damage? So then, even with .01 resistance, the potential energy of the charge still drops to 0 upon returning to the positive terminal of the battery? And finally, I also want to ask the same question I asked vofa - What happens if I add a component that requires 3.3 Volts and I need to figure out how to reduce the voltage to this utilizing a resistor? Let me know if other components are required. \$\endgroup\$ – seasidemango Jan 17 '18 at 19:34
  • \$\begingroup\$ @seasidemango If you know the exact resistance of the component that needs to be powered by 3.3V, you can use ohm's law (V=IR) to calculate such voltage drops. However, you chose to mention 3.3V, which is a common voltage for integrated circuits. Most integrated circuits do not have a fixed resistance, so you need a more complicated circuit to reduce the voltage. If you look up Voltage Regulators, you can learn more. \$\endgroup\$ – Cort Ammon Jan 17 '18 at 19:48
  • \$\begingroup\$ Gotcha, that makes a lot more sense now. I appreciate it! \$\endgroup\$ – seasidemango Jan 17 '18 at 19:55
0
\$\begingroup\$

Voltage does not flow. It is measured across something.

Current flows. It is measured through something. All current that flows out of one battery terminal flows back into the other terminal.

Ohm's law says

$$V_{battery} = I_{battery} * R_{load}$$ Rearranging the symbols: $$I_{battery} = \frac{V_{battery}}{R_{load}}$$

In your first example, $$R_{load} = 100\Omega, V_{battery}=5V$$ so $$I_{load} = \frac{5V}{100\Omega} = 50mA$$

In your second example, $$R_{load} = 50\Omega, V_{battery}=5V$$ so $$I_{load} = \frac{5V}{50\Omega} = 100mA$$

Battery current is determined by the load. Less resistance = more current. A short circuit would be when R gets very small, which makes I very big.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ This is well put, and I really appreciate the response! So the resistor determines the current output with a voltage source. What happens if I add a component that requires 3.3 Volts and I need to figure out how to reduce the voltage to this utilizing a resistor? Also, let me know if other components are required here. I am very new, but trying to get a base for all this. \$\endgroup\$ – seasidemango Jan 17 '18 at 19:23
  • \$\begingroup\$ @seasidemango: You would never use a resistor for it, since the voltage drop across a resistor depends on the current through it, and the current through it could be changing from moment to moment. \$\endgroup\$ – Ignacio Vazquez-Abrams Jan 17 '18 at 19:30
  • \$\begingroup\$ Interesting. If you don't mind me asking, what would you use? \$\endgroup\$ – seasidemango Jan 17 '18 at 19:35
  • \$\begingroup\$ @seasidemango You would likely use a linear regulator. It maintains a constant output voltage regardless of load current flowing through it--within its operating limits. (An example of such a device is LM1117-3.3- ti.com/lit/ds/symlink/lm1117.pdf). \$\endgroup\$ – vofa Jan 17 '18 at 19:48

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.