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What is the second/lower 15 k-ohm resistor for?

I understand why there is a 12 k-Ohm resistor in the top, that's for the V-out. But why is there a 15 k-Ohm resistor in the lower right? Why would you need to make a voltage drop on the path back to the battery, isn't that just waste of energy?

Voltage drops.

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    \$\begingroup\$ what will be the output voltage if you remove the 15 k Ohm resistor? \$\endgroup\$ – jsotola Dec 26 '19 at 0:05
  • \$\begingroup\$ If you remove the 15kΩ resistor (and replace it with a wire), then Vout is connected directly to the negative terminal of your battery (which you would probably define as 0V). I doubt that you want Vout to always be 0V. \$\endgroup\$ – Harry Dec 26 '19 at 0:15
  • \$\begingroup\$ @Harry why would it be 0V? \$\endgroup\$ – GoldenRetriever Dec 26 '19 at 8:23
  • \$\begingroup\$ If you define the positive terminal of the 9V battery as +9V, then the negative terminal of the battery must be 0V. If you connect Vout directly to 0V (with a wire) then the voltage you measure at Vout (relative to 0V) is 0V. \$\endgroup\$ – Harry Dec 26 '19 at 16:12
  • \$\begingroup\$ You can understand your whole circuit using one equation, Ohm's Law: V = IR. The current flowing in the circuit is I = V/R = 9V / (12kΩ + 15kΩ) = 0.333mA. Therefore, the voltage dropped across the 12kΩ resistor is V = IR = 0.333mA x 12kΩ = 4V. Similarly, the voltage dropped across the 15kΩ resistor is V = IR = 0.333mA x 15kΩ = 5V. \$\endgroup\$ – Harry Dec 26 '19 at 16:25
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What is the second/lower 15 k-ohm resistor for?

I understand why there is a 12 k-Ohm resistor in the top, that's for the V-out.

The arrangement shown gives \$ V_{OUT} = \frac {15}{12+15}9 = 5 \ \text V\$.

But why is there a 15 k-Ohm resistor in the lower right?

You wouldn't need it if the load required 5 V and was exactly 15 kΩ. This could be the case in some situations. In many cases you might not know the load impedance or it could change somewhat from device to device so it makes sense to make a divider using known resistor values to drive the following circuit.

Why would you need to make a voltage drop on the path back to the battery, isn't that just waste of energy?

  • If the 15 kΩ resistor was not there then VOUT would be 9 V as there is no current flowing so there is no voltage drop across the 12k resistor. When current is drawn from VOUT the voltage drop will vary with the current. This may not be desirable.
  • All current must return to the battery in any case.
  • Yes, energy is consumed and no, the circuit is (hopefully) performing a useful function so there is a benefit to the energy cost.

From the comments:

But why is Vout 9V without the 15 k-ohm resistor?

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. With a very high input impedance voltmeter (10 MΩ, say) so little current will be drawn that there will be no voltage drop across R1 so the meter will read 9 V.

I thought that the Vout is only dependent on the 12 k-ohm resistor, as Vout lies above the 15 k-ohm resistor, and current flows from left to right and then down and back to battery? What you are writing is that, the Vout is also dependent on the lower resistor, even though Vout lies just above it. How can that be?

Perhaps the easiest way to imagine this is to consider a potentiometer.

schematic

simulate this circuit

Figure 2. Various configurations of voltage divider.

  • In Figure 2a we have added a GND point from which all voltage measurements will be taken. We have combined your potential divider into one resistor, R1: 12k + 15k = 27k. It should be clear that the voltage reading at the bottom of R1 will be 0 V and at the top will be 9 V.
  • In Figure 2b we have replaced R1 with a potentiometer R2. I hope that it is clear that the voltages at the bottom and top are still 0 V and 9 V respectively. The wiper on the potentiometer is shown in mid-position. Q: What do you think the VM2 reading will be? A: 9 V / 2 = 4.5 V. As the wiper runs from bottom to top the VM2 reading will increase linearly from 0 to 9 V and will be 9 V times the ratio of the lower resistance divided by the total resistance. \$ V_{OUT} = \frac {R_L}{R_{Tot}} 9 \ \text V \$.
  • In Figure 2c we have replaced R2 with R3 and R4 which are in the ratio 12k:15k which is 4:5. This is equivalent to having the wiper of R2 at 5/9 of the way towards the top. From the previous paragraph we can see that this should give the result \$ V_{OUT} = \frac {R_L}{R_{Tot}} 9 = \frac {5}{9} 9 = 5 \ \text V \$.
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  • \$\begingroup\$ But why is Vout 9V without the 15 k-ohm resistor? I thought that the Vout is only dependent on the 12 k-ohm resistor, as Vout lies above the 15 k-ohm resistor, and current flows from left to right, out of Vout, and then down to the lower resistor to find its way back to battery. What you are writing is that Vout is also dependent on the lower resistor, even though Vout lies above it. How can that be? \$\endgroup\$ – GoldenRetriever Dec 26 '19 at 9:41
  • \$\begingroup\$ See the update. \$\endgroup\$ – Transistor Dec 26 '19 at 10:04
  • \$\begingroup\$ Thanks for the detailed description and the depictions. It's much appreciated. \$\endgroup\$ – GoldenRetriever Dec 26 '19 at 11:04
  • \$\begingroup\$ Thank you for accepting my answer. Have I managed to remove all confusion? \$\endgroup\$ – Transistor Dec 26 '19 at 11:15
  • \$\begingroup\$ indeed you have! \$\endgroup\$ – GoldenRetriever Dec 26 '19 at 20:16
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That is a voltage divider, one of the most basic building blocks in electronics. The Vout node is 5V in respect to battery negative terminal. If there was no lower resistor, Vout would be 9V because without current there is no voltage drop in a resistor.

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  • \$\begingroup\$ This is incorrect. If there were no lower resistor, then Vout would be 0V. The current through the remaining resistor would not be zero, it would be 9V / 12kΩ = 0.75mA. \$\endgroup\$ – Harry Dec 26 '19 at 0:11
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    \$\begingroup\$ My answer holds if the lower resistor is removed so it is an open circuit. Your comment holds if lower resistor is replaced by short circuit, which even you seem to doubt if that is what the OP wants. \$\endgroup\$ – Justme Dec 26 '19 at 0:18
  • \$\begingroup\$ Ah, thanks for the clarification. Your answer makes sense now. \$\endgroup\$ – Harry Dec 26 '19 at 0:24
  • \$\begingroup\$ @Justme, can you clarify a bit more? I don't understand why a resistor, that lies just under the Vout, could have an impact on what's going out on Vout? I thought that the 12 k-ohm resistor served that purpose, when flow goes from left to rigt, through the top resistor, through the lower resistor, and then finally back to battery again. \$\endgroup\$ – GoldenRetriever Dec 26 '19 at 10:02
  • \$\begingroup\$ Like you said, there is current flowing through the resistors. According to Ohm's law there is voltage over a resistor that has current flowing through it. So the Vout depends on the resistances of both resistors. Same reason you can have 10 lamps of 24V each in a series string to power them from 240V mains voltage. Each lamp has 24V over it. \$\endgroup\$ – Justme Dec 26 '19 at 11:07
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First of all, if your circuit itself does not have a Ground (0 V) reference, just probing a wire anywhere and finding voltages just isn't gonna work because there is no reference voltage to compare it to. That is why voltage across something is called Potential Difference.

Now assuming the negative terminal of the battery is connected to Ground and the Vout too is measured with respect to the same Ground, what you have is a simple Voltage Divider.

schematic

simulate this circuit – Schematic created using CircuitLab

So basically what you are measuring is the potential difference across the 15k resistor. Now if you didn't have that resistor and replace it with a short circuit, the current would pass unhinged (assuming short-circuit wire resistance being 0) and by Ohm's Law would result in a Zero voltage drop (V = IR = I x 0 = 0), which makes your Vout essentially zero as it is now directly connected to the Ground reference (i.e., the negative terminal of your battery).

schematic

simulate this circuit

Similarly, you can't live without the 12k too as then the battery would concentrate all its voltage across the 15k and since Vout measures across the 15k, you will get all of the 9V, defeating the purpose of the voltage division, implemented using the two resistors.

schematic

simulate this circuit

Both of them are necessary for this particular operation as it is the unequal division of the source voltage across each resistor that results in the desired required output.

Sure, if you need all of your source voltage, don't bother plugging in resistors (but don't short circuit your battery too). Cheers!

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  • \$\begingroup\$ I am completely new to electronics, but what is the ground for? \$\endgroup\$ – GoldenRetriever Dec 26 '19 at 19:45
  • \$\begingroup\$ A Ground is basically a reference point. When measuring a voltage from a "point" on a wire, or more aptly, from a node, the potential at that point is measured with respect to another point. That "another point" is made the ground. Measuring a voltage at a point on a wire is meaningless unless it is referenced to some other point, usually a zero volt potential, such that potential difference = V(at the point) - V(ground, which is taken zero) = V(at the point) \$\endgroup\$ – nn08 Dec 26 '19 at 19:50
  • \$\begingroup\$ It is meaningless because potential difference is defined as the difference between two potentials. A single potential by itself isn't enough. \$\endgroup\$ – nn08 Dec 26 '19 at 19:52
  • \$\begingroup\$ aha! I get it. Is that why there is always ground added to a circuit? \$\endgroup\$ – GoldenRetriever Dec 26 '19 at 19:59
  • \$\begingroup\$ Precisely. Are there any more confusions regarding this? \$\endgroup\$ – nn08 Dec 26 '19 at 20:00

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