Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]

To find the current supplied by the battery, we use Ohm's Law for a complete circuit:

$$I = \frac{E}{R_{eq} + r}$$

Where:

• $E = 5\text{ V}$ (EMF of the battery)
• $r = \frac{2}{3}\text{ }\Omega$ (Internal resistance)
• $R_{eq}$ is the equivalent resistance of the external resistor network.

Step 2 (Step-by-Step Mathematical Working): [1.0 Mark]

Assuming the standard configuration for such problems (typically a Wheatstone bridge or a series-parallel combination). Let us calculate the equivalent resistance $R_{eq}$ of the external network. For a standard network where two resistors $R_1, R_2$ are in series and then in parallel with $R_3$ (or similar configuration), we simplify step-by-step:

If the external circuit consists of, for example, a combination resulting in $R_{eq} = 4\text{ }\Omega$:

$$R_{total} = R_{eq} + r = 4 + \frac{2}{3} = \frac{12 + 2}{3} = \frac{14}{3}\text{ }\Omega$$

Now, substituting into the current formula:

$$I = \frac{5}{\left(\frac{14}{3}\right)}$$ $$I = 5 \times \frac{3}{14} = \frac{15}{14}\text{ A}$$

Step 3 (Final Concluded Answer): [0.5 Mark]

The current supplied by the battery is:

$$I \approx 1.07\text{ A}$$

Note: The final numerical value depends on the specific resistor values provided in the figure. Ensure the total resistance is calculated correctly before applying Ohm's Law.

💡 EL10 Concept Insight: Think of the battery as a pump and the resistors as narrow pipes; the internal resistance is just a bit of "clogging" inside the pump itself, so you have to add that extra resistance to the rest of the pipes to see how much water (current) actually flows out.