RRB Technician Grade I 13 March 2026 Paper

Concept:By Kirchhoff's voltage law (KVL), the source voltage equals the sum of the load voltage and the diode voltage at every instant. A pure sinusoidal source has zero average value over one complete cycle.Explanation:The diode voltage $v_D(t)$ is the voltage across the diode from anode to cathode, and $v_L(t)$ is the voltage across the resistive load.Applying KVL around the loop at any instant gives:$v_s(t) = v_L(t) + v_D(t)$Taking the average of both sides over one complete cycle:$\bar{v}_s = \bar{v}_L + \bar{v}_D$Since the source is a pure sinusoid, its average over one complete cycle is $\bar{v}_s = 0$.Therefore, $\bar{v}_L + \bar{v}_D = 0$, which gives:$\bar{v}_D = -\bar{v}_L$So the average values of $v_L(t)$ and $v_D(t)$ are equal in magnitude and opposite in sign.Reversing the diode orientation only changes which half-cycle conducts, so $\bar{v}_D$ is still nonzero; the rms value of $v_D(t)$ is $V_m/2$, not the source rms $V_m/\sqrt{2}$; and $\bar{v}_D$ is not zero because it equals $-\bar{v}_L$.Answer:Option B is the correct statement.