Calculating Error in Efficiency of Half Wave Rectifier

[STEMucator]

Homework Statement

A single phase half wave rectifier produces a DC signal $V_{DC} = (4.6 \pm 0.78) V$ and AC signal $V_{AC} = (0.22 \pm 0.036) V$. What is the error on the efficiency?

Homework Equations

Functional dependence, propagation of error:

$\sigma_f = \sqrt{ (\frac{\partial f}{\partial x})^2 \sigma_x^2 + (\frac{\partial f}{\partial y})^2 \sigma_y^2 + ... }$

Efficiency:

$\eta = \frac{0.405}{1 + \frac{R_{Gen}}{R_{Load}}}$

The Attempt at a Solution

I'm actually not quite sure where to start this. Does the efficiency have a functional dependence on the AC and DC voltage somehow?

$V_{DC} = I_{DC}R_{Load}$
$V_{AC} = I_{AC}R_{Gen}$

 

[STEMucator]

Hey guys, turns out there was a typo in the question. The error on the quality factor is what was required, not the efficiency.

The quality factor is given by: $Q = \frac{V_{DC} - V_{Ripple}}{V_{DC}}$

Applying error propagation:

$\sigma_Q = \sqrt{ (\frac{\partial Q}{\partial V_{DC}})^2 \sigma_{V_{DC}}^2 + (\frac{\partial Q}{\partial V_{Ripple}})^2 \sigma_{V_{Ripple}}^2 }$

This gives:

$\sigma_Q = \sqrt{ (\frac{V_{AC}}{V_{DC}^2})^2 \sigma_{V_{DC}}^2 + (\frac{1}{V_{DC}})^2 \sigma_{V_{AC}}^2 }$

Which yields the correct answer after subbing everything in.

Sorry for the confusion.