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The circuit.And why is there a negative 5V at the bottom? Provided image

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    \$\begingroup\$ There's -5V at the bottom because someone put it there. The rest is just a voltage divider. \$\endgroup\$ – Ignacio Vazquez-Abrams Oct 12 '17 at 2:24
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    \$\begingroup\$ It is a way to check if you understand what potential difference is ! \$\endgroup\$ – MatsK Oct 12 '17 at 3:59
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You are supposed to imagine the following in your head:

schematic

simulate this circuit – Schematic created using CircuitLab

That's where the ground reference comes from; in this case, the mid-point between the two voltage rails.

If you want a formulaic method for coming the voltage at the output then you can first compute the current through the two series resistors as \$I=\frac{\left(+5\:\textrm{V}\right) - \left(-5\:\textrm{V}\right)}{R_1+R_2}=\frac{10\:\textrm{V}}{5\:\Omega}\$, which equals \$2\:\textrm{A}\$. The voltage drop of \$R_1\$ is clearly \$2\:\Omega\cdot 2\:\textrm{A}=4\:\textrm{V}\$. Dropping \$4\:\textrm{V}\$ from the \$+5\:\textrm{V}\$ upper rail leaves you \$+1\:\textrm{V}\$ at the (+) terminal there. The (-) terminal, by definition, is \$0\:\textrm{V}\$. So the difference between them is, again, just \$1\:\textrm{V}\$.

You could also come up from the bottom by first computing the voltage drop across \$R_2\$ as \$3\:\Omega\cdot 2\:\textrm{A}=6\:\textrm{V}\$. Adding that value to the bottom rail voltage of \$-5\:\textrm{V}\$ gets you right back to \$+1\:\textrm{V}\$ at the (+) terminal.

Either way, same answer. Good thing.

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The bottom of the divider is at -5 V because the person who designed the question made it so. Pretty much by definition, a voltage divider divides the overall voltage present at the ends, whatever they are.

  1. With +/-5 V sources, the overall voltage is 10 V.
  2. Use Ohm's Law to determine the current through the string.
  3. Use Ohm's Law to determine v3.
  4. Because the two power sources are equal but opposite, the circuit Ground, 0 V, is half-way between the two supplies.
  5. Add v3 to the negative power potential to get the output voltage with respect to GND.
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  • \$\begingroup\$ I still don't get how to get 1V?I used ohm's law and I got 2A.V3 is 2×3=6VAnd then what?? \$\endgroup\$ – Sook Lim Oct 12 '17 at 5:55
  • \$\begingroup\$ @SookLim "5. Add v3 to the negative power potential to get the output voltage with respect to GND." and what's 6V + (-5V)? \$\endgroup\$ – Doodle Oct 12 '17 at 7:29
  • \$\begingroup\$ But why is 0V halfway between two supplies?Because one resistor is 2 ohm and another is 3 ohm \$\endgroup\$ – Sook Lim Oct 12 '17 at 8:42
  • \$\begingroup\$ A voltage always is the potential difference between two points. Both the +5 and -5 values are with respect to something. Also, there must be a return path for the current from each supply. Connecting the +5 V and -5 V terminals of two supplies to a circuit will not deliver any power if the return terminals of both supplies are left floating. By convention, the common connection for the two supplies is called Ground. The same would be true (but the answer would be different) if the two supplies were +8 V and -2 V. Ground still would be the 0 V reference potential for both supplies. \$\endgroup\$ – AnalogKid Oct 12 '17 at 11:47

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