Re: a-level question help Please!?


<paulreynolds245@xxxxxxxxx> wrote in message
news:1194250267.673039.64130@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
: Hi There
:
: I'm writing an a-level paper in about four days and I'm having
: problems finding info on a particular question in one of the past
: papers. None of my text books have the info I need to resolve the
: problem. The paper is available at the following link:-
:
: http://www.cie.org.uk/docs/dynamic/15856.pdf
:
: The mark scheme(answers for the paper) is available here:-
:
: http://www.cie.org.uk/docs/dynamic/15810.pdf
:
: The question I'm having difficulty with is Question:3b(i)1
:
: It asks for the mean voltage of a pd across a load resistor in an RC
: circuit. The circuit consists of a half-wave rectified ac source, the
: load resistor and a capacitor in parallel with the resistor. The
: resistor resistance is given, as well as a graph of of the pd/time for
: the rectified pd, also, a graph for the smoothed rectified pd.
:
: My difficulty is finding the mean voltage across the load resistance
: when connected with the capacitor. My guessing would be the it's the
: average value between the ripple max and min, but the exam paper's
: corresponding answer *** does not agree.
:
: Any help what so ever would be much appreciated.

Ok...

From fig 4.2 you can see the voltage as a function of time without
the capacitor installed, but you cannot trust the graphs for accuracy.

Let's look at 4.2 for a moment, and imagine that you did what
you are guessing. The average value between the max and min
is 3V, but look at the AREA between the times 0 and 0.03
and 6V. Clearly the area beneath the curve is NOT half the total
rectangular area.
If you found the area under the sine curve, and then flat-lined
that to make a rectangular area between t = 0 and t = 0.03,
then the height would give the mean voltage without the capacitor
and resistor. It will be
(Area under curve) / (total area) * 6V, which is less than 3V.

Now comes the tricky part, but to make it easy to understand
let's put in the capacitor and take out the resistor.
Because you have a diode in the circuit, current cannot flow
back the other way and the voltage across the capacitor has
to rise to 6V and stay there.
Replacing the resistor will bleed current away from the capacitor
and cause the voltage to fall, but it will top up again when the
signal generator again supplies a current to charge the capacitor,
and that is shown in fig 4.3.

Unfortunately I cannot find a value for the capacitance to calculate
an RC time constant
http://www.tpub.com/neets/book2/3d.htm
and so we are forced to estimate it from the graph, which looks
to be slightly less than 5.5 V midway between the peak voltage
of 6V and the minimum of about 5V.



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