Explanation: The instantaneous amplitude of a sine wave at a given phase angle ($\theta$) is calculated using the formula $V(\theta) = V_{peak} \times \sin(\theta)$. In this case: * The peak voltage ($V_{peak}$) is 5 volts. * The phase angle ($\theta$) is $\pi/3$ radians. First, calculate the sine of the angle: $\sin(\pi/3 \text{ radians}) = \sin(60^\circ) = \sqrt{3}/2 \approx 0.866$. Now, substitute this value into the formula: $V(\pi/3) = 5 \text{ V} \times 0.866$ $V(\pi/3) = 4.33 \text{ V}$ Therefore, the amplitude at $\pi/3$ radians is approximately +4.3 volts. Options A and B are incorrect due to the negative sign, as $\pi/3$ (or 60 degrees) is in the first quadrant where sine values are positive. Option C is incorrect as 2.5 volts would correspond to $\sin(\theta) = 0.5$, which is $\pi/6$ radians (30 degrees).