Does changing capacitance in an RLC circuit, while keeping the resistance and inductance constant, dampen the voltage across the resistor? While doing an experiment on RLC circuits, I noticed that by decreasing the capacitance, while keeping the resistance and inductance constant, the resonant frequency changes (which is to be expected), but the graph also changes shape by becoming wider and lower. Why does this happen?
Here is a graph of my results:(edit) After redoing my experiments base on Farcher's suggestions, I got the following results: 

And here is the circuit diagram of my experiment set-up (sorry about the rough sketch)

 A: You obviously have taken care in taking your observation and produced a really nice set of results which I have tried to analyse further for you.  
It definitely would have helped if you had measured the dc resistance of your inductor and also stated the values of the capacitors which you used in your experiment.  
Assuming that the inductance of your inductor was $18$ mH then from your graphs one can read off the resonance frequencies and using $f = \dfrac{1}{2 \pi \sqrt{L C}}$ find the value of the capacitance of the capacitors that you used.
At this stage it would have been useful to compare the experimental values found by experiment (using the given value of the inductance of the inductor) and the values written on the capacitors.
With the help of a drawing app the values which I got from your graphs with A as the set of data with the lowest resonant frequency etc was:
$\text{A} , 80 \text{Hz}, 210 \mu \text{F}; \text{B}, 190\text{Hz}, 37\mu \text{F}; \text{C}, 290\text{Hz}, 17 \mu \text{F}; \text{D}, 430\text{Hz},7.7\mu \text{F} ; \text{E},650 \text{Hz}, 3.3\mu \text{F} ; \text{F}, 880\text{Hz}, 1.8\mu \text{F}$
@CuriousOne has made some comments about your experiment and given you possible reasons why theory and experiment do not agree.
The link that he quotes predicts that for a series LCR circuit the $Q$ value of the circuit is given by $Q = \dfrac 1 R \sqrt {\dfrac L C}$.
So as the capacitance $C$ decreases the $Q$ value of the circuit should increase.
If this did happen it might be explained by the inductor becoming lossy due to eddy currents in the iron core.
The suggestion to increase the resistance of the resistor across which you have measured the voltage (and hence the current) in the circuit to $1000 \; \Omega$ would be an attempt to mask the increased losses in the inductor at higher frequencies but unfortunately would probably produce a very dull set of results in that the $Q$ value of the circuit would be reduced by a factor of approximately 100 and thus you would get a set of almost straight line curves.  
The quite unexpected and interesting thing that I have found by taking further measurements from your graphs is that the $Q$ value of you circuit does indeed increase as the capacitance in the circuit decreases.
To analyse your data more accurately one would need access to your readings but from the graphs that you provided it is possible to show the increase.
The Q-value is defined as the value of the resonant frequency divided by the half power band width.
As you have measured voltages the half power band width is the width of the response curve when the voltage is the voltage at resonance divided by the square root of two.  This is shown in the reference that you have been given.  
Taking some measurements from your graphs gives $Q$ vales as follows:
$\text{A} , 1.1; \text{B}, 1.4; \text{C}, 1.6; \text{D}, 1.6 ; \text{E},1.8; \text{F}, 1.7$
To illustrate the "optical illusion" I have rescaled graph A - the green one - which appears to have the highest Q value by expanding the frequency scale by a constant factor so that the resonant frequency is the same as that of graph D - the orange one - whilst at the same time scaling the voltages at resonance to be equal and superimposed the scaled graph A on your original set of graphs.

Note that the scaling that I have done to graph A does not change its $Q$ value.
To analyse things further you need to know the resistance of the inductor so that you have a value of the total resistance in the circuit - $R$ in the equation above for $Q$ value.  
First impressions are that theory and experiment do not agree and then the lossy nature of the inductor might be the cause.
Rather than increase the resistance in the circuit a way forward might be to remove the iron core from the inductor if that is possible, and repeat your experiment?
However this would not address the possibility of losses in the iron core.
Again looking at your results it is interesting to note that the peaks of all the response curves are not at the same value of voltage but decrease as the frequency increases.  
If the power supply gave a constant output voltage then at resonance the current should be the same and hence the voltage across the $10 \; \Omega$ resistance irrespective of the value of the capacitor because at resonance the current only depends on the supply voltage and the resistance of the circuit.
The fact that it drops as the frequency increases could either mean that the supply voltage drops with frequency of there are significant losses in the inductor.
One way to check what is happening is to make additional measurements of the voltage across the terminals of the power supply.  
I suspect that you will find that even with a constant voltage supply the peaks will be lower as the frequency increases due to the lossy inductor.
