Question

n resistors ,each of resistance R are first connected in series so as to give an effective resistance Rs.Subsequently these resistors are connected in parallel so as to give an effective resistance Rp .Then the ratio Rs/Rp will be *
2 points
n/1
n^2/1 (n square /1)
1/n^2(1/n square)
1/n

Answer 1

Answer:

Given:

n resistors are first connected in series to get net resistance as Rs. They they are connected in parallel to get net resistance as Rp

To find:

Ratio of Rs and Rp

Calculation:

First we have series combination :

$Rs = R + R + .....n \: times$

$= > Rs = nR$

Then we have parallel combination :

$\dfrac{1}{Rp} = \dfrac{1}{R} + \dfrac{1}{R} + .....n \: times$

$= > \dfrac{1}{Rp} = \dfrac{n}{R}$

$= > Rp = \dfrac{R}{n}$

Now required ratio will be :

$\boxed{ \red{ \large{ \bold{Rs : Rp = n : \dfrac{1}{n} = {n}^{2} : 1}}}}$

Answer 2

GiveN :

• The number of Resistors is n, and each is of R ohm.
• They are first connected in series to give effective resistance of Rs.
• Then they are connected in parallel to give effective resistance as Rs.

To FinD :

• Ratio of Rs and Rp

SolutioN :

First Connect them in series :

$\longrightarrow \sf{R_s \: = \: R_1 \: + \: R_2 \: + \: ...... \: + \: R_n} \\ \\ \longrightarrow \sf{R_s \: = \: R \: + \: R\: + \: ...... \: + \: R_n} \\ \\ \longrightarrow \sf{R_s \: = \: Rn \: \: \: \: \: \: ...(1)}$

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Now, Connect them in parallel :

$\longrightarrow \sf{\dfrac{1}{R_p} \: = \: \dfrac{1}{R_1} \: + \: \dfrac{1}{R_2} \: + \: ...... \: + \: \dfrac{1}{R_n}} \\ \\ \longrightarrow \sf{\dfrac{1}{R_p} \: = \: \dfrac{1}{R} \: + \: \dfrac{1}{R} \: + \: ..... \: + \: \dfrac{1}{R_n}} \\ \\ \longrightarrow \sf{\dfrac{1}{R_p} \: = \: \dfrac{n}{R} \: \: \: \: \: \: ...(2)}$

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Ratio is :

(2) ÷ (1)

$\longrightarrow \sf{\dfrac{R_s}{R_p} \: = \: \dfrac{Rn}{\dfrac{n}{R}}} \\ \\ \longrightarrow \sf{\dfrac{R_s}{R_p} \: = \: \dfrac{n^2}{1}} \\ \\ \underline{\underline{\sf{Ratio \: is \: n^2 \: : \: 1}}}$