An ideal half-wave rectifier supplies a purely resistive load from a sinusoidal source. Let v L (t) be the loa

Question

An ideal half-wave rectifier supplies a purely resistive load from a sinusoidal source. Let v L (t) be the load voltage and v D (t) be the diode voltage, measured with reference to the anode. Which of the following statements is always true over one complete cycle?

Examoogle · Accepted Answer

The correct answer is Option B: The average values of vL(t) and vD(t) are equal in magnitude and opposite in sign.. In a series circuit consisting of a source $v_s(t)$, a diode $v_D(t)$, and a load $v_L(t)$, Kirchhoff's Voltage Law (KVL) dictates that the sum of voltages must equal the source voltage at any instant. Over one complete cycle, the average value of the sinusoidal source voltage is zero, leading to the condition where the average values of the load and diode voltages must be equal in magnitude and opposite in sign.

Step-by-Step Solution & Concepts:
In a series circuit consisting of a source $v_s(t)$, a diode $v_D(t)$, and a load $v_L(t)$, Kirchhoff's Voltage Law (KVL) dictates that the sum of voltages must equal the source voltage at any instant. Over one complete cycle, the average value of the sinusoidal source voltage is zero, leading to the condition where the average values of the load and diode voltages must be equal in magnitude and opposite in sign.

Working Principle: According to KVL, $v_s(t) = v_D(t) + v_L(t)$. Taking the integral over one full period $T$, we get $\frac{1}{T} \int_{0}^{T} v_s(t) dt = \frac{1}{T} \int_{0}^{T} v_D(t) dt + \frac{1}{T} \int_{0}^{T} v_L(t) dt$. Since the average value of a pure sinusoidal source over a cycle is zero, it follows that $0 = V_{D(avg)} + V_{L(avg)}$, implying $V_{D(avg)} = -V_{L(avg)}$.

Key Formulas: $v_s(t) = v_D(t) + v_L(t)$ — Kirchhoff's Voltage Law for the loop; $V_{avg} = \frac{1}{T} \int_{0}^{T} v(t) dt$ — Definition of average value over period T

Key Points: The average value of a symmetric sinusoidal signal over a full cycle is zero.; For a half-wave rectifier, the load voltage $v_L(t)$ is positive during the conduction half-cycle.; The diode voltage $v_D(t)$ is zero during conduction and matches the source voltage during non-conduction, resulting in a negative average value.; KVL must hold instantaneously for every time $t$.

Applications: AC to DC power conversion; Signal rectification in electronics

Additional Info: Option D is incorrect because while the diode conducts for half a cycle, it blocks the other half where the full source voltage appears across it, resulting in a non-zero average.. Option C is incorrect because the RMS value of $v_D(t)$ is $V_m/2$ for a half-wave rectifier, whereas the source RMS is $V_m/\sqrt{2}$.

Answer: B is correct — By Kirchhoff's Voltage Law, the sum of average voltages around the loop must equal the average source voltage, which is zero for a sinusoid.

Exam Tip: Always remember that the sum of average values of voltage drops across components in any closed loop is zero, regardless of the circuit nonlinearity.

Examoogle · Answer

Option A: The average value of vD(t) becomes zero if the diode orientation is reversed.

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Option C: The rms value of vD(t) equals the rms value of the source voltage.

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Option D: The average value of vD(t) is zero because the diode conducts for half the cycle.

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