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I'm a British high school student (doing a certificate called A levels) and question 16 in this past paper (1st Link ) is baffling me. The answers to the questions are on the mark scheme (2nd Link). I want someone to please explain the scenario for me. Part (b)i) is the hard part. The circuit is shown in part (c) of the question. I've done 10 hours of solid work on this question, and I have an idea of what the answer may be, but I'm finding it hard to articulate my thoughts because I keep running into too many inconsistencies.

Here is what I would like to kindly ask an answer for:

  1. May someone please explain the graph describing the variation in the voltage of the D.C power supply?

  2. May someone please answer part (b)i) in a more coherent form than the mark scheme? (This is the most important question of all.) Qualitative analyses are preferable, but mathematical ones are welcome if that is how you think.

  3. May someone please explain how the graph in part (c) of the question arises?

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  • \$\begingroup\$ Here are some links that I found that put my thoughts into words. 1. forum.allaboutcircuits.com/threads/… (Check the 4th comment to the guy's question.) 2. uk.answers.yahoo.com/question/index?qid=20140603104026AAUi7JG (Look at the last comment in this link.) \$\endgroup\$ – Mathematician Oct 19 '16 at 4:42
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    \$\begingroup\$ Look up "bridge rectifier" for 16(a) or else "full-wave rectification" -- either the DC output of a bridge rectifier applied to single phase AC or else the DC output of two diodes applied to split-phase AC (found in the US, commonly.) Speaking of which, the frequency of the graph shown is from the North American grid system and not the UK. It's 120 Hz, not 100 Hz (full-wave doubles the number of DC pulses.) Are test writers too lazy in the UK? For the other questions look up "ripple voltage" and "DC power supply filter capacitors". \$\endgroup\$ – jonk Oct 19 '16 at 5:36
  • \$\begingroup\$ And how did you answer the question #6? It contains a big clue to your problem with Q#16 \$\endgroup\$ – Ale..chenski Oct 19 '16 at 5:57
  • \$\begingroup\$ Ali, I haven't done the whole paper. It was just question 16 that I looked at. \$\endgroup\$ – Mathematician Oct 19 '16 at 6:24
  • \$\begingroup\$ Jonk, I will look all that stuff up, I promise. However, may you please provide n explanation as well. This is just so that it skims stuff down for me. \$\endgroup\$ – Mathematician Oct 19 '16 at 6:26
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I agree with pjc50 and want to stress the severity of the problem. The exam test #16 says:

"Figure 1 shows the output from the terminals of a power supply labelled d.c. (direct current)."

This looks OK and is well recognizable to a seasoned engineer as an output from a bridge rectifier, with all implication of asymmetrical diodes and such. But how is a schoolchild supposed to know this? The "educators" failed to to tell.

I noticed that the concept of realistic source impedance is not taught/mentioned. Although the question (a) might be reasonably formulated (but still with ambiguity), without information on the nature of source impedance all following tasks are meaningless and confusing.

This exam text seems to be in circulation for a couple of years. I can imagine how much frustration and emotional distress was inflicted by this faulty test on young and especially mathematically-inclined minds.

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So, the question is:

A teacher suggests that certain electronic circuits require a constant voltage supply to operate correctly. (i) A student places a capacitor across the terminals of this power supply. Suggest how this produces a constant voltage.

And the marking scheme says

Capacitor stores charge/charges up

(If voltage is constant) capacitor doesn’t discharge

.. both of which are true, but as you say, not obviously related to voltage. Recall the definition of capacitance: $$C = {dQ\over dV}$$

So whenever the capacitor is confronted with a change in voltage, it responds by changing its charge. The capacitor counteracts the change in voltage. When the input voltage is rising: "Capacitor stores charge/charges up" applies. When the input voltage is falling: "(If voltage is not constant) capacitor does discharge" applies. The capacitor holds up the voltage while discharging through the load.

What is not shown is that the input must contain a diode or similar component, so if the input voltage is lower than the capacitor plate voltage then the capacitor does not discharge back into the power supply.

(I'm 20 years past A-levels and still find the marking schemes obtuse, they're simplified beyond the point of understanding)

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enter image description here

1) The Signal is wavy but has an average positive value so it consistent with the label “d.c voltage”

2) Assuming it is a DC power supply and not an output from a bridged audio power amp, adding a capacitor will store the positive peak voltage while conducting high current when the supply exceeds the capacitor voltage and thus holds the peak as a steady d.c voltage

3) the circuit added appears to be a resistor current to ground which discharges the capacitor voltage between positive input peak charges

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  • \$\begingroup\$ The expected answer to 16a is probably something like "it doesn't switch direction". \$\endgroup\$ – immibis Jul 4 '18 at 1:01
  • \$\begingroup\$ Yes something like that \$\endgroup\$ – Sunnyskyguy EE75 Jul 4 '18 at 1:36
  • \$\begingroup\$ Equally important in 1) is that the voltage never drops below zero. \$\endgroup\$ – WhatRoughBeast Nov 10 '18 at 18:37

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