A resistor is connected to a battery of 12 V emf and internal resistance 2 Ω. If the current in the circuit is 0.6 A, the terminal voltage of the battery is: ____.

We are given:

• Electromotive force (emf) of the battery, $E\; =\; 12 V$.
• Internal resistance of the battery, $r\; =\; 2 \Omega$.
• Current flowing in the circuit, $I\; =\; 0.6 A$.

The terminal voltage $V\_t$ of a battery is the voltage that appears across its external terminals when a current $I$ is drawn. It is reduced from the emf by the voltage drop across the internal resistance:

$V\_t\; =\; E\; -\; I\; \backslash ,\; r$

Substituting the known values:

$V\_t\; =\; 12 \backslash text\{V\}\; -\; (0.6 \backslash text\{A\})(2 \Omega )$

Calculate the product $I\; r$:

$0.6 \backslash text\{A\}\; \backslash times\; 2 \Omega \; =\; 1.2 \backslash text\{V\}$

Now subtract this drop from the emf:

$V\_t\; =\; 12 \backslash text\{V\}\; -\; 1.2 \backslash text\{V\}\; =\; 10.8 \backslash text\{V\}$

Therefore, the terminal voltage of the battery is 10.8 V, which matches the provided correct option.