# Help With Calculations

Discussion in 'General Electronics Discussion' started by nabila hussain, May 4, 2017.

1. ### nabila hussain

2
0
May 4, 2017
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

2. ### Muki

2
0
Oct 6, 2016

5,165
1,087
Dec 18, 2013
Yes I agree Muki. Nice to see you on here at long last

4. ### nabila hussain

2
0
May 4, 2017
Hi,

No information given, its a unit that I have picked up from a previous teacher who has now left and no example work, so a bit lost with it.

5,165
1,087
Dec 18, 2013
What school are you working at?
Thanks

6. ### BobK

7,682
1,688
Jan 5, 2010
Your calculation looks right up until you compute the current through the combined resistance. You used 9V as the voltage across the equivalent resistor. But it is not 9V, some of the voltage is across the LED. Use the forward voltage of the LED, subtract that from your 9V then do the calculation again. It will come out closer to correct.

It may still not be correct, because the forward voltage across the LED varies with current and varies from one unit to another. The forward voltage given as a specification for the LED is a typical value at a specific current. If you do not know the forward voltage of the LED use 2V for red, yellow, or green, 3.3V for white or blue.

Bob

7. ### duke37

5,364
770
Jan 9, 2011
Resistors in parallel.
How do you get the right answer when using the wrong equation?
As Bob says, the driving voltage is the battery voltage minus the led voltage. This is given near the bottom of the table.
Calculate the current from the total resistance with this voltage.
Calculate the voltage across R4 subtract this from the previous voltage to give the voltage across R1, R2, R3.
Calculate the currents through R1, R2, R3.

5.8mA is not a lot different to 5.3mA and could be due to voltage, temperature or resistor accuracy.

8. ### Audioguru

3,144
698
Sep 24, 2016
Look again. The schematic shows R1 as 100 ohms but your calculation wrongly has it as 200 ohms.

9. ### FuZZ1L0G1C

368
118
Mar 25, 2014
There are 3 major errors with tabled equations supplied:
The 1/R resistor inversions should preferably be separated by parentheses or at least the entire 'divisor' group of resistors.
I prefer individually bracketing each resistor as some hand-held calculators ('in the field') get cnofused.
Adding these inverted values results in a very low value so to correct this the entire group is inverted (hence the grouped resistor answer being the divisor.
Using Excel syntax (also works on scientific calculator, the correct equation is Rp=1/((R1)+(R2)+(R3)) so
=1/((1/200)+(1/150)+(1/220)).
A quick-and-nasty method of proving that your Rp answer falls within the correct range (although not accurate)
is to average the values then as 3 'virtual' values are equal divide by 3. (average(R1,R2,R3,...))/qty
=(average(200,150,220))/3
Currents along the series parts of the circuit are equal.
On parallel components the remaining 2 R's 'shunt' part of the current away from R being tested.
This is why in-circuit measurements differ from ohm's law I=V/R.
Voltage-drop is proportional to resistance (a higher resistance between points A and B drops voltage more at B).
1490R measured for 1500R (R4) looks like a close-tolerance 1% resistor.
Error%=((1500/1490)-1)*100
I created a spreadsheet>PDF which has all calculations including R1 as 100R and 200R, plus examples of non-parentheses and non-inversion errors in parallel (resistors and inductors) or series capacitor formulae.
https://www.electronicspoint.com/at...4/?temp_hash=213d15c388fac2de48ae5bfa9ba9aa46

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367.6 KB
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