# Can the current through this $6 \text{ }\Omega$ resistor be zero? [closed]

In other words, can the $$6 \text{ }\Omega$$ resistor below have $$I = 0$$? I think the answer is yes, depending on the direction of the current.

Also, what do the negative $$\epsilon$$ values in the circuit signify? I think they signify which way the current is flowing?

• Can you clarify where you get negative $\epsilon$ from? Is that part of the problem specification? As a general hint, just add up the currents and voltages according to Kirchoff's laws, and then assume that the current through the middle wire is $0$. That will give you a couple of linear equations that let you solve for $\epsilon$. Commented Mar 20, 2021 at 18:15

This comment and this other answer propose two long routes for the solution. Let me propose you a shorter one.

Denote by A and B the two nodes of your circuit. When $$I=0$$, it is indeed $$V_\mathrm{AB} = \mathcal{E}$$. Furthermore, when $$I=0$$, the branch with $$\mathcal{E}$$ and the $$6\,\Omega$$ resistor can be removed without changing $$V_\mathrm{AB}$$, and the voltage divider's formula yields

$$\mathcal{E}=V_\mathrm{AB} = \frac{3\,\Omega}{2\,\Omega+3\,\Omega}\times 28\,\mathrm{V} = 16.8\,\mathrm{V}.$$

Yes, an easy way to see this is to use source transformation. Starting with your original circuit:

First, I'll transform your 28V source and series 2$$\Omega$$ resistor into a current source with parallel resistor:

Next, I'll combine the 2$$\Omega$$ and 3$$\Omega$$ resistors which are now in parallel (results in 6/5 $$\Omega$$ resistor):

Finally, transform back into a voltage source and you will see exactly what value of E your middle branch battery must be to result in no current in the branch of interest:

Using simple circuit analysis tools (source transformation, superposition etc.) can often greatly simplify the problem at hand.

I like @MassimoOrtolano's answer as the simplest presented for your particular case.