# Will current flow if there's no return path?

Here is the problem I was trying to solve:

Find the potential difference between the points A and D

I used Kirchhoff's voltage law for the left loop and right loop and found out the current through the left loop to be $\frac{10}{2+3}$ A (2A) and for the right loop $\frac{20}{4+6}$ A (2A), both flowing clockwise. But this does not take into account the current between B and C (The connecting wire)? By book says current will not flow through BC and they proceeded to find the potential difference by adding\subtracting the potential drops along the way while taking current through that wire as 0.

One explanation was that it's because there's no return path for the current. But even during Earthing, there's no return path, yet charges flow for a short while.

My question: Why does current not flow through the BC path? If there exists a potential difference between B and C of 4V, charges should flow, right? Shouldn't all the current eventually pass only theough the loop at a lower potential?

Edit: What about a case like this?

Will current flow now?

• There will be 4 volts difference between B and C. No current will flow (after stabilization) along the link between B and C, because there will be no voltage differential across the resistor. – Hot Licks Dec 10 '17 at 2:01
• What do you mean by "even during earthing, there's no return path"? – The Photon Dec 10 '17 at 2:02
• (Better asked here: electronics.stackexchange.com) – Hot Licks Dec 10 '17 at 2:02
• The sentence as written doesn't make sense, but generally the term "earthing" means to provide a return path through the earth, so if I do try to make sense out of it, it's wrong. – The Photon Dec 10 '17 at 2:03
• Between B and the positive terminal of the 4V battery there is zero volts potential. Across the battery is 4V potential. From the negative terminal to C is zero volts. That adds up to 4V. No voltage across the resistor. – Hot Licks Dec 10 '17 at 2:28

If there exists a potential difference between B and C of 4V, charges should flow, right?

No, if there were a current through the 1 ohm resistor, the voltage $V_{CB}$ could not be $4\,\mathrm{V}$. This result is an elementary application of KVL and Ohm's law:

$$V_{CB} = I_{CB}\cdot 1\Omega + 4\,\mathrm{V}$$

See that only in the case that $I_{CB} = 0$ is $V_{CB} = 4\,\mathrm{V}$

• What if that 1Ω resistor wasn't there in that BC link? Will current flow then? (Assuming everything to be ideal) Something like this: i.stack.imgur.com/r9s4Z.png – Rick Dec 10 '17 at 5:47
• @Rick, no even replacing the $1\Omega$ resistor with a wire will not allow current $I_{BC}$ through the $4\,\mathrm{V}$ source. Perhaps the easiest way to see this is add a resistor of resistance $R$ between nodes $B$ and $D$ and solve for $I_{BC}$ as a function of $R$. Then find that $I_{BC} \rightarrow 0$ to zero as $R \rightarrow \infty$. – Alfred Centauri Dec 10 '17 at 13:36
• Oh ok got it, that step of adding that extra R didn't really strike at first – Rick Dec 10 '17 at 13:48

If there were a constant current between B and C, then the left and the right part of the circuit would charge indefinitely with the charges of opposite signs, the potential difference between the left and right parts would increase and eventually the current between B and C will stop.

• You mean the accumulated opposite charges will create a new potential difference causing them to flow back, right? – Rick Dec 10 '17 at 3:16
• @Rick: I mean causing the current between B and C to stop. If you need more details, you may read something on the transient processes of capacitor charging, say electrical4u.com/transient-behavior-of-capacitor. I mean every circuit has some capacitance (which can be very small). – akhmeteli Dec 10 '17 at 3:26