If n resistors connected series then after connected parallel what is the effective resistance ratio
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If they are in series we can just simply add them up (and I know you know why).
Let resistance of each resistor be R.
Now let's muster up some physics courage and solve the question.
If there are N resistors , we have to add R , N times I.e.
R+R+R......+R (N times)
Again , I know that you know that repeated addition is called multiplication.
So we can simply write it as
N×R=NR ..........(eq 1)
Coming to next step (the parallel one)
Now this is slight tricky. But as always I know that you know that we take the reciprocal of resistances and add them up if they are in parallel.(Why is it so, is another topic of discussion but I can tell you if you want).
What we have to do is take reciprocal of R and add N times I.e.
1/R +1/R +1/R.......+1/R (N times)
or simply multiply it
which is N×1/R = N/R..........(eq 2)
Now take the ratio by dividing eq1 by eq2
I.e. NR/(N/R)
= R^2 (R squared)
Hope that, it does put a smile om your face.
-----THANOS
Let resistance of each resistor be R.
Now let's muster up some physics courage and solve the question.
If there are N resistors , we have to add R , N times I.e.
R+R+R......+R (N times)
Again , I know that you know that repeated addition is called multiplication.
So we can simply write it as
N×R=NR ..........(eq 1)
Coming to next step (the parallel one)
Now this is slight tricky. But as always I know that you know that we take the reciprocal of resistances and add them up if they are in parallel.(Why is it so, is another topic of discussion but I can tell you if you want).
What we have to do is take reciprocal of R and add N times I.e.
1/R +1/R +1/R.......+1/R (N times)
or simply multiply it
which is N×1/R = N/R..........(eq 2)
Now take the ratio by dividing eq1 by eq2
I.e. NR/(N/R)
= R^2 (R squared)
Hope that, it does put a smile om your face.
-----THANOS
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