How to Calculate Electronics Formulas and Compare to Measured Values in Class?

Hi All,

I am a teacher and covering a unit on electronics and finding some of the calculations a little difficult, electronics is not my strong point!

I would really appreciate it if some one is able to use formulas to find the answers with the working outs. I am not sure what I am doing wrong but I am getting different answers to the measured values.

Any help with be greatly appreciated.

Thanks

Nabila

Work book

Hi,

I get an error when trying to download the document. Also, perhaps provide as a text document or other format that may exclude viruses etc.

I am using Linux - so perhaps LibreOffice format ?

Regards,

Richard.

Hi Richard,

Thank you for looking in to it, i'm not sure how to turn it into LibreOffice, would PDF work?
Or google doc link?

Hi Nabila,

Thanks - i had to accept the site security certificate in the browser to download.

You parallel resistor calculation has a 200ohm resisitor - this should be 100ohms.

A quick look - the calculation method is ok for the resistors (despite error for 200ohms should be 100ohm resistor), but the diode will have some internal impedance/resistance. As such, the voltage across the resistive components should be used to calculate the current through those resistive components.

That is - to calculate I, you need to change the formula to :

I = (9 - LED Voltage Drop) / 1547

Despite you error in the text, your result for the parallel resistors is correct

The ammeter may have a slight impact on this accuracy too - as it will have an internal impedance.

With the voltmeter you can measure voltages across the resistors easily (i assume a digital voltmeter), and it is a simple calculation.

In stating this - are you looking to create a formula to calculate the current through the resistances as part of teaching people how to do it ?

So, the voltage across the parallel resistors will be :

Volt_Parallel = (9volts - LED Voltage Drop) * (Parallel_Res / (Parallel_Res + 1.5k) )

Does the above help ?

To add - i can see that you measured the voltage across the diode (LED) - can you confirm that the battery is actually 9volts ? Thanks.

The reason is, that if you add your measured voltages for R4, R2 and D1, they add up to 11.31volts - and not 9volts.

Regards,

Richard.

Hi Richard,

I plan to able to show them the working out as i did on the first section current in circuit. Or even understand first, myself how the answers were calculated, in a step by step format.

I used a power supply and set the voltage to 9v, I may have to do the measurements again, i used a digital volt meter.

Thanks for the help so far, I am going to attempt the answers with the guidance and see if I am able to get correct values. If you have done any, would you send them to so i can see?

Thank you
Nabila

Hi Nabila,

OK – to go through the calculations as per the worksheet :

Rp = 1 / (1/R1 + 1/R2 + 1/R3 )

Rt = (Rp + R4)

I = (V_B1 – V_D1) / Rt

Current through R3 (if we base it on I)

Ir3 = (I * Rp) / R3

Current through R1 (if we base it on I)

Ir1 = (I * Rp) / R1

Voltage drop across R4 (if we base it on I)

V_R4 = I * R4

Voltage drop across R2 (if we base it on I)

V_R2 = (I * Rp) = V_R1 = V_R3

Since resistors R1, R2 and R3 are in parallel.

Voltage drop across D1, since this is an active device and we do not have the datasheet, then we use the following based on I :

V_D1 = (V_B1 – (I*Rp) – (I*R4)) = (V_B1 – I*(Rp + R4)) = (V_B1 - I*Rt)

The key aspect is that V_B1 and measured V_D1 is accurate from the start.

To calculate the resistance of R4 then it is :

R4 = V_R4 / I

Hope the above is understandable.

Regards,

Richard

Thank you Richard for your support, greatly appreciated.

Have a lovely weekend.

Hi,
Richard has given you a great set of answers.
However I though I might offer the following observations of my own problems in setting up a similar practical session some 50 years ago in a teaching facility.

I got caught out when I set this practical up for 20 students to do simultaneously. The students complained that the measured answer differed from the expected results and that no student had an exactly similar result.

I had forgotten about the tolerance of the resistors I was issuing to the class (between 5% and 10%).

If your tech has time and the resources then it is probably a good idea to match each resistor value used so that they are as near to each other as possible. In the end I had a sets of matched resistors in lots of 20 so that I could bask in the glory of Students getting the result they were expecting.

I got out of the class room problem on that day, by usurping the teachers lecture and suggesting that the students pooled results and looked at the bell curve of their results and that they also worked out the possible tolerance of the resistors they were using.