Question

The maximum dc output power in a Half wave rectifier occurs when the load resistance is \___ the diode forward resistance.

• A. Same as
• B. Double
• C. Half
• D. Quadruple

Hint:

The Maximum Power Transfer Theorem states that maximum power is delivered to a load when the load resistance equals the Thevenin (source) resistance.
For a rectifier, consider the diode's forward resistance as part of the source resistance when the diode is conducting.

Answer 1

The Correct Option is A

This question relates to the condition for maximum power transfer. For a source with an internal resistance \(R_s\) (in this case, the diode forward resistance \(r_f\)) to deliver maximum power to a load resistance \(R_L\), the load resistance must be equal to the source resistance. \[ R_L = R_s \] In the context of a half-wave rectifier, the diode (when conducting) has a forward resistance \(r_f\). The rest of the circuit (transformer secondary, etc.) can be considered part of the source resistance. If we are considering only the diode's forward resistance as the "source resistance" seen by the load \(R_L\), then for maximum DC output power to be delivered to \(R_L\), we need \(R_L = r_f\). The question is specifically about "maximum dc output power". The efficiency of a half-wave rectifier is given by \(\eta = \frac{P_{dc}}{P_{ac,input}}\). Maximum efficiency is about 40.6% when \(R_L \gg r_f\). However, for maximum *power delivered to the load* from a source with internal resistance, the matching condition applies. Let \(V_s\) be the RMS voltage of the AC source feeding the rectifier and diode. The diode acts as a switch with forward resistance \(r_f\). The DC current \(I_{dc} = I_m/\pi\), and DC voltage \(V_{dc} = I_{dc}R_L\). Peak current \(I_m = V_m / (r_f + R_L)\), where \(V_m\) is peak AC voltage. DC power \(P_{dc} = I_{dc}^2 R_L = (I_m/\pi)^2 R_L = \frac{V_m^2}{\pi^2 (r_f+R_L)^2} R_L\). To find when \(P_{dc}\) is maximum with respect to \(R_L\), we differentiate \(P_{dc}\) with respect to \(R_L\) and set it to zero. Let \(K = V_m^2/\pi^2\). \(P_{dc} = K \frac{R_L}{(r_f+R_L)^2}\). \(\frac{dP_{dc}}{dR_L} = K \frac{(r_f+R_L)^2 \cdot 1 - R_L \cdot 2(r_f+R_L) \cdot 1}{(r_f+R_L)^4} = 0\). \((r_f+R_L)^2 - 2R_L(r_f+R_L) = 0\). Since \(r_f+R_L \neq 0\), divide by \(r_f+R_L\): \(r_f+R_L - 2R_L = 0\) \(r_f - R_L = 0 \Rightarrow R_L = r_f\). Thus, maximum DC output power occurs when the load resistance \(R_L\) is the same as the diode forward resistance \(r_f\). \[ \boxed{\text{Same as}} \]