A set of n identical resistors each of resistance R are connected in series and the effective resistance is found to be X. When these n resistors are connected in parallel the effective resistance is found to be Y find the ratio of X and Y
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In series connection :
effective resistance of n resistors is given by Rs = R + R + R + R ....... = nR = X .........(1)
In parallel combination :
effective resistance of n resistors is given by 1/Rp = Y = 1/R + 1/R + 1/R + 1/R + ........
Rp = R/n = Y ..........(2)
dividing eq1 by eq. 2, we get :
X/Y = n2
effective resistance of n resistors is given by Rs = R + R + R + R ....... = nR = X .........(1)
In parallel combination :
effective resistance of n resistors is given by 1/Rp = Y = 1/R + 1/R + 1/R + 1/R + ........
Rp = R/n = Y ..........(2)
dividing eq1 by eq. 2, we get :
X/Y = n2
uttu94:
it's 4marks bro not 2 or 3
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In series
nR = X
The connection in parallel
R/n = Y
X : Y = nR divided by R/n
R and R will be cancelled
And we get n^2
X / Y = n^2 / 1
So,
X : Y = n^2 : 1
Hope it helps you!!!
nR = X
The connection in parallel
R/n = Y
X : Y = nR divided by R/n
R and R will be cancelled
And we get n^2
X / Y = n^2 / 1
So,
X : Y = n^2 : 1
Hope it helps you!!!
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