Determine the currents flowing through the three resistors depicted...

Based on the given circuit diagram, we can analyze the flow of current step-by-step:

1. Identify the short circuit: There is a plain horizontal wire connected directly across the upper $4\ \Omega$ resistor. This wire has zero resistance and acts as a short circuit. Current always follows the path of least resistance, so all the current will flow through this plain wire, bypassing the upper $4\ \Omega$ resistor entirely. Therefore, the current through the upper $4\ \Omega$ resistor is $0\text{ A}$.

2. Simplify the circuit: With the upper resistor shorted out, the effective circuit is a single closed loop containing the left $4\ \Omega$ resistor, the right $6\ \Omega$ resistor, and the two batteries at the bottom.

3. Determine the total resistance: The remaining two resistors are in series.
Total resistance, $R_{\text{eq}} = 4\ \Omega + 6\ \Omega = 10\ \Omega$.

4. Determine the net electromotive force (EMF): Observe the polarities of the two batteries. The longer vertical line represents the positive terminal, and the shorter, thicker line represents the negative terminal.
For both the $4\text{ V}$ and $2\text{ V}$ batteries, the positive terminal is on the left. They are connected such that the negative terminal of the $4\text{ V}$ battery is connected to the positive terminal of the $2\text{ V}$ battery. This means they are in a series-aiding configuration.
Net EMF, $V_{\text{net}} = 4\text{ V} + 2\text{ V} = 6\text{ V}$.

5. Calculate the current: Using Ohm's law for the simplified loop:
$I = \dfrac{V_{\text{net}}}{R_{\text{eq}}} = \dfrac{6\text{ V}}{10\ \Omega} = 0.6\text{ A}$.

This current of $0.6\text{ A}$ flows through both the left $4\ \Omega$ resistor and the right $6\ \Omega$ resistor.

Thus, the current is $0\text{ A}$ in the upper $4\ \Omega$ resistor and $0.6\text{ A}$ in the remaining two resistors.

Answer: $0\text{ A}$ in the upper $4\ \Omega$ resistor and $0.6\text{ A}$ in the remaining two