GCSE Electronics

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Question 4 offers an easy two marks providing you can remember your op-amp stuff:

Non-inverting input


Don't let them confuse you with the + and - signs here. Although they are used for positive and ground, in the case of an op-amp or comparator they refer to the inputs. Look carefully whenever you see the triangular symbol. + is non-inverting, and - is inverting. Again the examiners are not your friends and they just can't wait to trip you up.


Question 5 wants to know what you know about resistors in series and in parallel:

Arrangement C

Arrangement A

Let's break those arrangements down into labelled parts:

Two in series

One alone

Two in parallel

A handy tip to remeber here is that resistors arranged in paralell as in arrangement C always give a total resistance smaller than the smallest resistor in it. That way you can answer this sort of question at a glance. Another one is that if both resistors are the same then the value of the total resistance is half that of either resistor: or as these are 10k the value would 5k.


Doing it the mathy way we'd simply add together the resistors in series. They're both 10k, and adding them together gives us 20k. For resistors in parallel however, we have to add the reciprocals to get the reciprocal of the total resistance. Or more simply:


1/ Rt  = 1/R1 + 1/R2      so    1/ Rt   = 1/1000 + 1/10,000  and from that 1/Rt = 0.0002    and to get the value of the resistance in total we just shift the the 1/ to the other side  like this:  Rt = 1/0.002 and if you do that on your calculator you get 5000,  or 5k.


They haven't done with you yet. Now they want to know if you can identify resistors by the colour of their stripes.




The data sheet will help you here. Remember that the first two stripes are the numeral. Brown is 1, black is 0. The next one along is the multiplier, and that's three. You multiply by powers of ten, and 10 x 10 X 10 = 1000 (the easy way is to say there's three zeros). So 10 X 1000 = 10,000 or 10k.


Now they want you to show you know the principle of an unloaded voltage divider. The question runs:

(15 / 15+10) 15........(15/25)15.........0.6 (15) = 9V

Once again the data sheet will come to your rescue here because it gives you the formula you are supposed to use as above. Remember that R1 is attached to the positive and R2 is attached to the negative and Bob's your Uncle.

So, you tot up the value of the two resistors, and divide R2 by what you get. Then you multiply it by the supplied voltage to get the voltage out. Piece of cake. But there is another way you can do it in your head.

Tot up the resistors. In this case you get 25k .......ignore the k as they are both in the kilohm range and call it 25. The bottom one is three-fifths of the total value, or 0.6 in decimals. 0.6 x 15 is 9.....9V.


Want some proof? Here it is: