Step 1: Read the condition. The current through the $8\,\Omega$ resistor must be the same before and after the extra source $E$ is connected. If the current does not change, the voltage across that resistor cannot change either (Ohm's law). Step 2: Equivalent resistance before $E$. Before connecting $E$, the $12\,\text{V}$ source drives the series chain $6\,\Omega$, $8\,\Omega$, $10\,\Omega$: \[ R_{\text{eq}} = 6 + 8 + 10 = 24\,\Omega. \] Step 3: Current in the loop. \[ I = \frac{12}{24} = 0.5\,\text{A}. \] Being a series loop, the $8\,\Omega$ resistor carries this same $0.5\,\text{A}$. Step 4: Voltage across the $8\,\Omega$ resistor. \[ V_8 = I\times 8 = 0.5 \times 8 = 4\,\text{V}. \] Step 5: Apply the unchanged-current rule. For the current through $8\,\Omega$ to stay the same after $E$ is added, the voltage across it must remain $4\,\text{V}$. Step 6: Identify $E$. The source $E$ placed across those terminals must therefore equal that voltage, $E = 4\,\text{V}$.
\[ \boxed{E = 4\ \text{V}} \]
