Question

if n equal resistance are connected in seriesand and then connected in parallel.what is the ratio maximum to minimum​

Answer 1

Answer:

Supposing that n resistors with equal resistance are connected in series.

So, the net resistance in series = R1 + R2 + .... + Rn

Now, R-series = n*R

We can write in such a way because all the resistors have equal resistance.

Now, latest suppose the same number of resistors with equal resistance are connected in parallel.

So, the net resistance in parallel = 1/R1 + 1/R2 + .... + 1/Rn

Now, 1/R-parallel = n*1/R

Hence, R-parallel = R/n

At last, the ratio between maximum resistance to minimum resistance will be ratio between the the net resistance in series to the net resistance in parallel.

Hence, n*R ÷ R/n

Solving further, (n*R)*n/R

So, n² is the final answer.

Answer 2

$\huge{\red{\underline {\underline {Question??}}}}$

If n equal resistance are connected in seriesand and then connected in parallel.What is the ratio maximum to minimum?

$\huge{\red{\underline {\underline {Answer:-}}}}$

Let the resistance of each resistor be R.

Series connection :

Equivalent resistance (maximum) Rs=

R+R+....+n terms

⟹ Rs =nR

Parallel connection :

Equivalent resistance (minimum)

$\frac{1}{rp} = \frac{1 }{r} + \frac{1}{r} + ......n \: terms$

We get ,

$= > \frac{1}{rp} = \frac{n}{r}$

$= > rp = \frac{r}{n}$

Thus ,ration of maximum to minimum, resistance

$= > \frac{rs}{rp} = \frac{nr}{r \div \: n}$

$= > \frac{rs}{rp} = {n}^{2}$

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