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According to the professor, adding the ground at 0V has no effect on the current flow and current flows in this circuit as if the grounding point did not exist so the answer is E.

How have people died from touching powerlines?

enter image description here

JRE
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Ardalan G.
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    Because a power line does not exist in isolation. – Tom Carpenter Jan 28 '21 at 22:51
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    Because they weren't touching a point that was at 0V. – hobbs Jan 28 '21 at 23:03
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    Possible duplicate of https://electronics.stackexchange.com/questions/529744/how-to-measure-if-there-is-current-between-point-a-and-b-when-source-and-groundi – MarkU Jan 28 '21 at 23:24
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    Possible duplicate of https://electronics.stackexchange.com/questions/53214/what-is-the-right-way-to-understand-the-behaviour-of-shocking-voltage – MarkU Jan 28 '21 at 23:24
  • Those are related but not the same. – Voltage Spike Jan 29 '21 at 00:20
  • you don't need math to know that E is correct. If the one ground is only one, it doesn't matter at all where it goes. – dandavis Jan 29 '21 at 05:47
  • Currents A and C must be the same due to conservation of charge. The battery moves charge but does not create or destroy it. A and B must be the same because it is just a wire between them. C and D must be the same because it is just a wire between them. Therefore E is the only answer that could be correct. – user57037 Jan 30 '21 at 22:29
  • See I would get this wrong because I would have assumed that the 12 ohm resistor gets shorted out. And I would have fought it until my face was blue. – Passerby Nov 30 '21 at 06:18
  • For your case, ground is not a part of the "circuit", not a circuit until.. – jay Dec 03 '21 at 20:20

5 Answers5

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In your example, the battery and resistors are floating as a separate circuit, except for the single tie to ground at one point between the two resistors. Because there is only that one ground tie, no loop to ground is formed, so no current flows into or out of ground. It’s the ‘loop’ part - or lack thereof - that your professor forgot to tell you about.

Anyway, the question is only interested in current. Because of this, we also know we can ignore the ground tie (no loop, so no current flows to or from it). It's a red herring.

This in mind, we can conclude that the only current flowing in the circuit is the loop formed by the battery and 2 resistors.

Kirchhoff's Current Law tells us that in a loop like that, the current is the same at all the points in the loop.

So the only sensible answer is (E), because all the currents are equal.


On the other hand, you raised the question about touching overhead power lines. Unlike the battery in the question, these are referenced to ground. Because of this, if you are unlucky enough to touch a downed line, and also are grounded, you complete a current path from the line, through your body, to ground, and then via ground back to the utility. And quite possibly, you die.

On the other, other hand, crows can land on power lines and survive just fine. Why? They practically mock you when they do that. That's just what crows do.

Crows, plotting... to dive-bomb your car and crap on it.

"Insulation is my super power." - Crow (photo credit: author)

Oh, you meant, why don’t they die?

Crows, just sitting there on the wire, doing their murderous crow thing, don’t complete that path to ground that would turn them into smoldering corvid cinders: they only touch the wire, but are otherwise insulated by the surrounding air. No current flows through their insouciant bodies, even though they are at the wire's high voltage potential.

There are limits. Birds, even those naughty, naughty crows, will avoid sitting on very high-voltage lines because as they approach the energized wire, the ensuing corona discharge causes them discomfort or even kills them. Crows have this figured out, and probably even teach each other about them. Not only that, there is evidence that birds and mammals can see corona discharge in UV. That’s avian nightmare fuel by any standard.

What's the birdie no-go voltage threshold? Hard to say. A quick look gives values of ~21kV or more (for reference, the lines shown in the photo above are probably 12kV.) This will vary with altitude and weather as these affect the air insulation strength. The size of the bird matters too: bigger birds have more surface-to-air coupling and thus would see more discharge effects.

As it so happens, humans can do this high-voltage party trick too, with a bit of help from a helicopter and a Faraday cage suit (so take that, crow!):

enter image description here

From this video: https://youtu.be/9YmFHAFYwmY (World Channel, https://www.youtube.com/channel/UCp7jpKjIOLFA1j3atWNJAKA)

The takeaway: no current flows to ground if there isn't a complete path that forms a loop back to the power source. Or, you're a crow.

hacktastical
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    Haha thanks for the informative and fun answer – Ardalan G. Jan 30 '21 at 04:56
  • In that video why is it arcing to the helicopter even though it's isolated? – Ardalan G. Jan 30 '21 at 05:04
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    The helicopter, although insulated by the air, still has some capacitance to ground. It’s a big object so that capacitance is non-trivial. The arc forms because the 500KV line is charging and discharging that capacitance. – hacktastical Jan 30 '21 at 06:06
  • @hacktastical I always wondered about this. I thought perhaps it was due to triboelectric effect of the helicopter blades. Because it doesn't really make that much sense that the helicopter would arc to the wire but the wire would not arc to the other wire, for example. – user57037 Jan 30 '21 at 22:31
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    @hacktastical K = Kelvin. k = kilo. – winny Jan 31 '21 at 09:39
  • Blame autocorrect. That’s it... – hacktastical Jan 31 '21 at 11:36
  • In a practical circuit, there may be some current when first connecting the floating circuit to ground in order to adjust the balance of static charges. But assuming that the circuit is already in a stable state, there is no current. – le_top Oct 26 '21 at 11:22
  • That wouldn’t apply to the question. It does apply to the helicopter, less so, that physics-taunting crow. – hacktastical Oct 26 '21 at 15:15
  • I always thought that if I touched the + side of a AA battery to the wall, it will discharge it. Now I am guessing that is not true. – Carlos Garcia Dec 03 '21 at 21:12
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    @CarlosGarcia, the amount that the battery would discharge depends on the loop formed by the (+) contact to the wall, through the floor, to your feet, through your body, and back to the (-) contact. This resistance is huge, the, the battery voltage low, so practically no current will flow. – hacktastical Dec 03 '21 at 21:19
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schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. In the case of a grid power line one of the wires has been neutralised by bonding it to an earth rod buried in the ground.

As shown in Figure 1, there are now two possible return paths to the supply transformer. One is through the neutral cable. The other is through the ground.

The important point to answer your question is that this circuit has two ground connections. Your question circuit only has one so there is nowhere that current can flow through that ground connection.

Transistor
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3

In your circuit, and in most electronic circuits, the Ground symbol just marks the point in the circuit that we will consider "Zero Volts" and will use as a reference when measuring voltages elsewhere in the circuit. The Ground symbol does not imply a connection to the earth, and has no effect on the current flowing in the circuit.

In AC power distribution, one wire (the Neutral) is connected to the earth, so touching the other ("Hot") wire and any conductor connected to the earth may result in a shock.

Peter Bennett
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  • And in that circuit, even if the grounding point _were_ connected to a metal rod driven into the ground, the results would be the same -- E is still the correct answer. – TimWescott Nov 30 '21 at 00:55
2

Need of understanding

In your place, I would ask the professor, "And what is the point of this circuit? Where is it used?" At least that is how I asked my professors in the break between lectures when I was a student... and the result was stunning - they started running away from me in panic at the end of the lectures and my exams were guaranteed with the highest grade:)

Of course, it was a joke (although there was some truth in it)... but what I want to say is that, in addition to clicking on questions, there must be understanding... and the basic circuits are best understood when their applications are shown. This way you will be able to use what is understood now in these simple electrical circuits in tomorrow's more complex electronic circuits. I will demonstrate it with your circuit.

Basic idea

Your circuit has a valuable application in electronic circuits - to convert a floating voltage to two single-ended voltages. To show it, I have visualized the voltages by voltage bars (in red) with proportional heights and the current by a loop (in green) - Fig. 1. I have assumed that R1 = R2, which is the most commonly used case in practice. As you can see, the input voltage V is split in two equal but with an opposite polarity output voltages V/2 and -V/2.

Question visualized

Fig. 1. OP's circuit visualized by voltage bars (in red) and a current loop (in green).

Actually, the two resistors form the well-known voltage divider... but what is unusual here is that its "output" (the middle point between the resistors) is grounded... so it is not an output... Instead, the other ends of the resistors are used as outputs.

Applications

1. MOSFET differential amplifier. Usually, differential amplifiers are driven by two single-ended input voltages applied to their inverting and non-inverting inputs. But what do we do when we have only one floating input voltage source? Then, your circuit can help us - Fig. 2.

MOSFET floating input

Fig. 2. The input part of a MOSFET differential amplifier driven by a floating input voltage source by the help of the OP's circuit.

Here is the full circuit of the MOSFET differential amplifier that is discussed in another related question.

MOSFET differential pair

Fig. 3. The full circuit of the MOSFET differential amplifier.

2. BJT differential amplifier. In a similar way, a BJT differential amplifier can be driven by a floating input voltage source that is split in two single-ended "voltage sources" - Fig. 4. This technique is discussed in a related RG question.

BJT floating input

Fig. 4. BJT differential amplifier driven by a floating input voltage source by the help of the OP's circuit (a snapshot of the whiteboard during a "brainstorming session" with my students).

In the applications above, the input floating voltage is converted to two single-ended output voltages with opposite polarities. But there are applications where we want single-ended voltages with the same polarity. How do we solve this problem?

3. Curve tracer. A typical application of this version of the OP's circuit is the simple curve tracer in Fig. 5 (you can learn more about it in another related RG question).

Curve tracer

Fig. 5. A circuit diagram of the simplest possible curve tracer for 2-terminal components like resistors, diodes, LEDs, etc. (a snapshot of the blackboard during the building circuit with my students).

As you can see, the input AC voltage is floating and is converted to two grounded voltage drops:

  • The first is the diode forward voltage VF across the diode that is applied to the first scope's channel Ch 1. It has the correct positive polarity.

  • The second is the voltage drop VR = IF.R that is applied to the second scope's input Ch 2 (ordinary oscilloscopes measure voltages; so the role of R is to convert the current to voltage). It directly represents the forward current IF through the diode since the value of R resistance is 1 kOhm ([V]/[kOhm] = [mA] -> [V] = [mA]). However, VR has an opposite polarity to VF... and the diode IV curve will be plotted in the fourth instead in the first quadrant. Here the manufacturers of oscilloscopes help us by putting an INVERT button on the faceplate. So press it and the image will appear in the first quadrant.

This unique circuit is another example of the Don Lancaster's "elegant simplicity". If we assume that the input voltage source and the scope are external devices, then this curve tracer consists only of two components - the diode under test and the resistor (Fig. 6).

Curve tracer on the board

Fig. 6. The curve tracer mounted on the prototyping board.

But the diode can be also considered as an external component; so this "curve tracer" is simply a resistor...


I hope that with my answer I have managed to awaken in you the desire to understand circuits and not just to formally answer your "clicker questions"...

Circuit fantasist
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Current always flows in a circuit. (Well almost always, "static" electricity being the exception.) That is, it returns to its source. All of the current from the battery in your schematic returns to its source, the battery, so none of it flows into the ground.

People die from touching high voltage power lines because they form part of a circuit. The circuit often involves the ground, metal appliances, water etc. But there is a circuit. Birds are able to sit on high voltage lines because they are not part of a circuit.

Math Keeps Me Busy
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    "(Well almost always, "static" electricity being the exception.)". Wrong, static electricity is no exception. – Vinzent Jan 28 '21 at 23:24
  • @Vinzent "static" electricity forms a circuit? – Math Keeps Me Busy Jan 28 '21 at 23:28
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    @Math, yup. It's capacitors. – Transistor Jan 28 '21 at 23:29
  • @Math, Yes it does. When you build up static electricity you are moving electrons (or positive charges) from one place to another. Think of it like lifting a heavy object against the force of gravity, and then dropping it (discharging).. – Vinzent Jan 28 '21 at 23:30
  • @Transistor, yes it is capacitors. – Math Keeps Me Busy Jan 28 '21 at 23:30
  • @Vinzent "When you build up static electricity you are moving electrons (or positive charges) from one place to another." Yup. – Math Keeps Me Busy Jan 28 '21 at 23:34
  • Static buildup deserves a footnote and separate explanation. I mean, how can a person get charged up walking across a carpet or rubbing against the upholstery of a chair? What is the circuit, exactly? Electrons are literally rubbing off and causing the person to take on net charge. And then when the person's net charge becomes balanced (in other words, when the ESD event occurs) what is the circuit? I think capacitance to ground can explain or model that. But the buildup is, well, odd. – user57037 Jan 30 '21 at 22:36
  • @mkeith Technically, Vinzent and Transistor are correct. The circuit consists of two parts, conductive current, which is the flow of charges, and displacement current which is a change in electric fields. We say that current flows through a capacitor, but the conductive current flows up to the dielectric, then displacement current flows across the dielectric, and then conductive current flows from the other plate around the rest of the circuit. One of Maxwell's laws, and Kirchhoff's Current Law. But we often think of "current" merely as flow of charges. Otherwise discharge of a capacitor ... – Math Keeps Me Busy Jan 30 '21 at 22:52
  • @mkeith ... via leakage through dielectric involves zero current. Not our intuition. See [this](https://electronics.meta.stackexchange.com/questions/7331/are-questions-of-the-form-what-is-the-best-way-to-explain-concept-phenomenon-x) for more info. – Math Keeps Me Busy Jan 30 '21 at 22:55
  • Yes. ESD is similar to capacitive discharge. I get that. And ESD guns (for lab testing ESD) ARE capacitive. But the process of a human body acquiring a net charge does not seem to be model-able that way. It is electromechanical. Mechanical friction is causing charges to transfer from one object to another. All I am saying is that it deserves a footnote and separate explanation. It isn't really like other circuits. – user57037 Jan 30 '21 at 23:03
  • The point is, change in electric field _IS_ a kind of current, at least to physicists, and at least in the sense that KCL is strictly true across capacitors. So when electrons leave a rug and enter my body (or the other way around) there is conductive current flow from the rug to my body, and an equal displacement current from my body to the rug. Thus charged, when I walk into another room, there is conductive flow (me moving while charged) and an equal displacement current, the electric field moving from one room to another. Intuitively seems like double accounting. – Math Keeps Me Busy Jan 30 '21 at 23:08