Hi, This is my data of Irms and angular frequency, and I have to plot the graph and teacher told me that the gradient will be capacitor's capacitance? i run the experiment with unknown capacitor and sinusoidal voltage across the capacitor is Vpeak = 2V and frequency varies. then teacher said to plot the graph from following data and he also said that the gradient will be capacitor's capacitance . He said to prove your method use known capacitor which is 0.2 micro F and we can do the same experiement on this capacitor as well and the gradient of graph will be capacitance of capacitor but i got wrong answer.. so am i doing correctly? and this is my excel result
You are varying the frequency and measuring the current. So the frequency is the independent variable (which should be displayed on the X axis) and the current is the dependent variable (which should be displayed on the Y axis). Try changing the axes of your chart and then try again. Be careful that you get the units correct as well. Then ask yourself, "why does the current vary with the frequency?" See if you can work out the formula.
Another way of seeing your error is what are the units of capacitance C? C = Q/V i.e. charge/voltage = current x time / voltage
UNITS! UNITS! UNITS! my friend. Capacitance has units of farads, or coulombs/volt. Your slope (gradient) has units of Y/X, or frequency/current or, in your case, 1/mAs, which is 1/mC. So you need to fold in voltage somehow -- and don't forget to be consistent and not mix peak and rms values together.
Your lab worksheet is correct if you look at Frequency [Hz] against current [mA]. But when you create the column with ω [radians/sec] you have managed to switch the two axes used for the chart. You still need to check the units (amps or milli amps) and convert from peak voltage to rms voltage. Did you work out the equation yet?
Right. That means plot rms current on the y-axis and angular frequency on the x-axis. You still have to do the suggestions stated above. 1) What is the formula for C? 2) What are the units of C? Doing the above will reveal your error.
1) Now convert that in terms of current, frequency and voltage. 2) From (1) write out the units in terms of current, time and voltage. The equation of a straight line is y = mx + c where y = dependent variable x = independent variable m = slope c = intercept, i.e. y at x = 0 The units of m must be the same as the units of y/x.
You need to find an equation that contains the data that you measure ( and frequency) and the capacitance that you require. Here is some help with the first few steps, have a go and post your efforts... Q = CV we know this equation, but we cannot measure Q (the charge). We can measure the current (rate of change of charge). Can you find an expression for a voltage that varies with a frequency ω and has a peak amplitude ? Can you differentiate this expression to find ?
okay it says i have use this formula but it is not giving me correct answer and i m using data of my spreadsheet https://www.dropbox.com/s/0thxwsrk3n8wl4w/lab.xlsx the worksheet is "2" Thanks
With the numbers that you've given (worksheet 2), I get C=48.3nF But this is not the same capacitor as used in worksheet 3. Look at the value of Worksheet 2 Worksheet 3 For similar values of V and f, the current differs by a factor of 10. Somehow you have got your measured data or your capacitors mixed up - only you can sort that one out! In the meantime, here is an explanation of what I can see in your spreadsheets. You have got the formula: Worksheet 2 is asking you to fix the frequency and plot against . The gradient gives you ωC, and you already know ω - so you can calculate C. Worksheet 3 is asking you to fix the voltage and plot against ω. The gradient gives you , and you already know - so you can calculate C. Hope that helps.
This is my circuit diagram. and i have to verify that my capacitor should be 0.1uF but i got wrong capacitance. Check page 2 of this forum. http://www.physicsforums.com/showthread.php?p=4382526&posted=1#post4382526
Looking at your spreadsheet, this may be the problem. A sine wave with a peak to peak voltage of 2V has an amplitude of 1V. The rms voltage is therefore or 0.71 V.