# Possibility of current flowing through this circuit [closed]

I am having some problems in solving this circuit. I will share my ideas here. Please correct my steps which are incorrect.

1. We emit current only from battery $$V_1$$ and let that current be $$I$$.

2)Now $$I$$ gets divided into $$I_1$$ and $$I_2$$ at point $$A$$ where $$I_1$$ flows through $$AC$$ and $$I_2$$ flows through $$AB$$.

3)At point $$D$$, $$I_1$$ further gets divided into $$I_3$$ and $$I_4$$ with $$I_3$$ going through $$DB$$ and $$I_4$$ going into the third loop. Now,i am in a great confusion. 1st of all from step $$3$$,the current returning to the battery is $$I_2+I_3 but we know that the same amount of current should return to the battery but here return amount is less than $$I$$.

2nd of all,the $$I_4$$ current which went into third loop,it goes and at $$E$$,it sees 2 junctions, so if $$I_4$$ further divides,then at a point we shall see two currents of opposite direction colliding each other,which cannot be possible.

Where am i making the mistake?Apparently the calculations are really easy once we assign appropriate currents,but here i am struggling with distributing currents.

• Hint: You have three separate simple circuits!
– Ed V
May 1 at 14:38
• Hint: No current flows from D to E. If current flows from D to E, there is no path for it to return. This violates Kirchoff Current Law... May 1 at 15:34
• @madness : You have not realized that $I_4$ is zero! May 1 at 23:41
• Why do you say $I_2+I_3<I$? That's really where you first got in trouble. Is it because $I_2+I_3+I=I$? If so, your original statement is only true if $I_4 > 0$. But you haven't proven that. In fact, you will find it false -- $I_4 = 0$. May 2 at 1:53
• The question is now closed, so there won’t be any more posted answers beyond my correct answer and the other two. Unfortunately, trick questions are sometimes assigned as homework or are even given on exams. I never did that when I taught my department’s entry level electronics class for first year chemistry grad students and senior undergrads of various majors. It is basically a bit cruel, but some people do it anyway. Here is another trick question over at the electrical engineering stack exchange: electronics.stackexchange.com/q/84447/223146.
– Ed V
May 2 at 11:00

Scroll down to see the experimental verification of my answer.

As in the hint I gave initially, this is three separate simple circuits. It is a trick question, such as might appear on a homework assignment or exam. The re-drawn diagram is:

There is no current flow in the ideal wire between points D and E: direct current cannot go both ways through that wire. So the wire between D and E does nothing at all and can be clipped, resulting in:

Points A and B are at the same potential due to the ideal wire connecting them. So shrink that wire to a point:

Now pull that point apart:

This is the same as the two previous diagrams: points A and B are still the same potential. Now clip the new wire between A and B, since there is no current flow through it:

Result: three simple loops with the obvious currents around them.

This is a trivially simple circuit to make and test, so I did. First, I built the OP’s circuit, using a standard breadboard, three fresh 9 V batteries, five 1% precision metal film resistors of 10 k ohms each, and a DMM in voltmeter and ammeter modes, as necessary. The circuit is shown here:

Sorry about the alligator clips. (I hate alligator clips.) Here is a closer view with annotation:

This is exactly as per the OP’s circuit. The individually measured resistances are shown at upper right. All three batteries measured 9.58 V. The measured currents were 0.46 mA, 0.95 mA and 0.47 mA, left to right.

Now the next figure shows the final state as three separate simple loops:

I re-did the current measurements and got the same values. These are shown in the figure.

• This is wrong: "There is no current flow in the ideal wire between points D and E: direct current cannot go both ways through that wire. So the wire between D and E does nothing at all." May 1 at 23:33
• Sorry, I don't have the lifespan to build every possible configuration of the circuit in order to exhaustively verify a negative. That's why we have theory. This is a theoretical question anyway. Set $V_1 = 0$, so that the left-hand part of the circuit doesn't contribute. Let $V_2$ and $V_3$ have the same potential between the terminals, but with both of $V_2$'s terminals at a higher potential/voltage than both of $V_3$'s terminals. Edit: remove $V_1$ from circuit, rather than setting its potentials to $0$. Just for simplicity. May 1 at 23:39
• @EdV this is a good answer and you are correct. The downvotes are very unfortunate +1 on my end
– Dale
May 2 at 0:31
• @Myridium the fact that there is no current in wire DE can be seen as follows: the current going down through R5 must be equal to the current going up through V3, and the current going up through V3 must be equal to the current going left through R4. Since the current going down through R5 is equal to the current left through R4, there is no current remaining to go through the wire DE. This answer is correct
– Dale
May 2 at 0:42
• @Myridium said “You are assuming that the batteries maintain a constant total charge”. Absolutely, I definitely assumed that. It is one of the three foundational assumptions of circuit theory. If you are not making that assumption then you are not doing circuit theory
– Dale
May 2 at 2:53

I guess I'd have a couple of questions before trying to respond.

1. When you state, "We emit current only from battery V1" are you suggesting that batteries V2 and V3 are essentially dead batteries? If so, they will have internal resistance which will provide the only current path through that region of the circuit, which will be high enough to limit current through those branches of the circuit to a level that is effectively negligible.
2. If all batteries are providing a charge, the voltage of each (are they equal?) and the amount of resistance in each resistor would determine current flow.

Since this is a DC (direct current) circuit: apart from a transient spike in current when the circuit is first assembled, the batteries will more-or-less retain a constant total charge (i.e. they won't gain an electrostatic charge). This is because any imbalances in the battery potentials will quickly equalise-- when electrons accumulate on a battery terminal due to current flow, it reduces the voltage of that terminal due to the electrostatic charge, much like a capacitor.

For this reason, the voltages on the battery terminals quickly equilibrate in such a way that the total electric charge contained in the batteries remains constant (this probably happens within nanoseconds or less; I don't know).

Therefore, you may assume that the current entering each battery is equal to the current which leaves. Using the conservation of current (Kirchoff's current law) at connection points between wires, you will quickly find that the $$DE$$ current must be zero, which means the right-hand loop can be treated as a separate circuit from the rest. Likewise, you can show that the remaining two loops more-or-less separate into their own circuits. ($$BA$$ carries the total current from both loops, but otherwise the two loops don't affect each other).