The terminal voltage of the battery, whose emf is 10V and internal resistance 1W, when connected through an external resistance of 4W as shown in the figure.

The circuit shown in the image consists of a battery with an electromotive force (emf) of $E\; =\; 10 V$ and an internal resistance of $r\; =\; 1 \Omega$. The battery is connected to an external resistor of $R\; =\; 4 \Omega$. To find the terminal voltage across the external resistor we first determine the current flowing in the circuit.

All resistances are in series, so the total resistance is

$R\_\{\backslash text\{total\}\}\; =\; r\; +\; R\; =\; 1 \Omega \; +\; 4 \Omega \; =\; 5 \Omega$

The current supplied by the battery is given by Ohm’s law:

$I\; =\; \backslash frac\{E\}\{R\_\{\backslash text\{total\}\}\}\; =\; \backslash frac\{10 \backslash text\{V\}\}\{5 \Omega \}\; =\; 2 \backslash text\{A\}$

The terminal voltage (the voltage you can measure across the battery’s external terminals) is the voltage drop across the external resistor, which is

$V\_\{\backslash text\{terminal\}\}\; =\; I\; \backslash times\; R\; =\; 2 \backslash text\{A\}\; \backslash times\; 4 \Omega \; =\; 8 \backslash text\{V\}$

Alternatively, the terminal voltage can be found by subtracting the internal‑resistance drop from the emf:

$V\_\{\backslash text\{terminal\}\}\; =\; E\; -\; I\; r\; =\; 10 \backslash text\{V\}\; -\; (2 \backslash text\{A\})(1 \Omega )\; =\; 8 \backslash text\{V\}$

Thus, the terminal voltage of the battery under the given load is 8 V, which matches the correct option provided.