1
$\begingroup$

enter image description here

Here $R1 = 2 \Omega$, $R2= 4 \Omega$ and $R3= 4 \Omega$

Though there looks to be a short circuit in this diagram, my teachers say that this circuit can easily be redrawn into simple parallel circuit. As far as I know a short circuit is an alternate way for current to move where there is no resistance whatsoever. So all of the current will flow through a short circuit if found.

Here if the current flows through that part of the wire ACB then after that R2 and R3 being the same it will get confused which way to go and even if it goes both sides then some part of the current is bouncing back and moving from the negative side of the cell to the positive. This is just a conjecture.

So can anyone please describe why there is no short circuit in the circuit though it seems to be? And also how to understand by looking at any circuit that there is actually no short circuit though there seems to be one.

$\endgroup$
2
$\begingroup$

As far as I know a short circuit is an alternate way for current to move where there is no resistance whatsoever. So all of the current will flow through a short circuit if found.

This is correct. The part that seems to be confusing you is that this isn't a short circuit.

As stated in the other answer, you have shorted the circuit between A and B using wire C; it will pass around R1 without issue.

Once you reach B, the current is no longer free to move without encountering a resistance. This means that the whole circuit is not shorted.

Comparing it to a circuit without wire C, you could say that C has shorted R1 and R2; but the total circuit has some resistance and is therefore not a short circuit.

It's only a short circuit if it can travel through the entire loop without encountering any resistance.

I believe this may also depend on the terminology you choose. It's very common to say you've made a "short circuit" between A and B; but I can see the logic behind saying the entire system itself is not a short circuit.

$\endgroup$
  • $\begingroup$ Alright I got a new concept. Can u help me by saying if it's right or not? When the current reaches point A, it would divide into two parts (because the circuit is not fully shorted), and the part that flows through R1 would eventually take the short circuit that is across R2 and R3 So neither of that part of current flow will flow through R2 and R3. But the part of flow which would take the wire C, would face two resistors on both sides and will divide into two parts and the part that flows through R2 will also take that short circuit across R2 and R3, and the other one would normally go. $\endgroup$ – Sami Jul 20 '17 at 12:17
  • $\begingroup$ There is no short circuit here in the sense that that all the resistors will have current flowing through them. See Farcher's response. You can trace from each side of each resistor to opposite ends of the battery. $\endgroup$ – Peter Hansen Jul 20 '17 at 22:24
2
$\begingroup$

It is best to label all the nodes.

enter image description here

$D$ and $F$ are in fact one node and are on one side of each resistor.
$A, \, B$ and $C$ are in fact one node and connected to the other side of each resistor so the three resistors are in parallel.

All the wires are doing is ensuring certain parts of the circuit are at the same potential.
If wire $C$ was not there you would only have current flowing through resistor $R1$ and likewise if wire $DF$ was absent current would only flow through resistor $R3$.
In both cases the remaining wire would be a short circuit across two of the resistors.
As drawn currents flow through all the resistors.

$\endgroup$
0
$\begingroup$

The wire which joins the points $A$ and $B$ is essentially a short circuit. Assuming the wire to be ideal (no resistance), the electric potential at point $A$ is equal to the electric potential at point $B$.

Therefore, the circuit is equivalent to the circuit where $R_2$ and $R_3$ were connected to point $A$.

With similar reasoning, you can do the same for the rest of the short circuit wires and obtain a simple parallel resistors combination.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.