# How to evaluate current in a circuit involving short circuit?

I am trying to solve this homework question: I got the short circuit in the middle by removing a voltage source. The problem is to find the current flowing through R6.

What I tried:

• Applying Star-Delta transformation on R3, R4 and R5 however it did not lead me anywhere.
• I also cannot seem to be able to apply mesh analysis as no voltages are supplied to me.
• I cannot convert the current source to a voltage source as there is no resistance in parallel with it.

Can someone please give me a hint as to how I should reduce this circuit?

Edit: Circuit after removing the short-circuited element R2:

Thanks to @Transistor and @MCG: I now was able to apply star-delta Thanks to @Transistor and @MCG: I now was able to apply star-delta transformation which reduced the circuit to this: I was able to solve this using current division.

P.S. The answer supplied by LTspice is 6 A for R6.

• Hint #1: One component is short-circuited so it can be removed. Redraw and post the modified circuit at the end of your question. Aug 15, 2018 at 9:06
• @Transistor Hi! I tried doing that, but the simulation answer changed from 6A to 5A in both LTspice and CircuitLabs, so I thought that was an incorrect operation. Is it actually valid? Aug 15, 2018 at 9:08
• You tried doing what? What component did you remove? Aug 15, 2018 at 9:09
• I'm sorry, you meant removing the 15ohm resistor. I misunderstood. Aug 15, 2018 at 9:10
• There are four 15 Ω resistors. Use the component numbers to be clear. Aug 15, 2018 at 9:11

You've worked it out on your own. Congratulations. Here's how I would have approached it. simulate this circuit – Schematic created using CircuitLab

Figure 1. The original. Note that R2 is short-circuited. It can be removed. simulate this circuit

Figure 2. Note that R3 and R4 are in parallel. simulate this circuit

Figure 3. Now we can add the series resistances. simulate this circuit

Figure 4. This is as far as we need to go.

You can apply your current division here.