Question

n equal resistors are first connected in series and then in parallel. The the ratio of equivalent resistance for series and parallel combination will be:

Answer 1

Answer :

n identical resistors are first connected in series and then in parallel.

We have to find the ratio of equivalent resistance for series and parallel connection$.$

Let resistance of each resistor be R.

◈ Equivalent of series :

➝ R = R1 + R2 + ... + Rn

➝ R = R + R + ... + R

➝ R = nR

◈ Equivalent of parallel :

➝ 1/R' = 1/R1 + 1/R2 + ... + 1/Rn

➝ 1/R' = 1/R + 1/R + ... + 1/R

➝ 1/R' = n/R

➝ R' = R/n

◈ Ratio of series to parallel :

➝ R/R' = nR/(R/n)

➝ R/R' = nR × n/R

➝ R : R' = n² : 1

☃ God_Bless :D

Answer 2

Answer :

• n² : 1 .

Explanation :

Given that,

• n equal resistors are first connected in series and then in parallel

to find,

• The ratio of equivalent resistance for series and parallel combination

so,

Let, resistance of each resistor be R

then,

By the law of combination of resistors in series :

stated as , The combined resistance of any number of resistances connected in series is equal to the sum of the individual resistances.

• R(eq) = R₁ + R₂ + R₃ + R₄ + ........      

The equivalent resistance in the circuit when n resistors are connected in series will be,

→  R₁(eq) = n R  ______equation (1)

     

And,

By the law of combination of resistors in parallel :

stated as, The reciprocal of the combined resistance of a number of resistors connected in parallel is equal to the sum of the reciprocals of all the individual resistances.

• 1/R(eq) = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄ + ......

The equivalent resistance in the circuit when n resistors are connected in parallel will be,

→ 1/R₂(eq) = n ( 1/R )  

→ R₂(eq) = R / n   _______equation (2)

so,

the ratio of equivalent resistance for series and parallel combination will be

→ R₁(eq) / R₂(eq) = ( n R ) / ( R / n )

→ R₁(eq) / R₂(eq) = n² : 1

therefore,

• Ratio of equivalent resistance for series and parallel combination is n² : 1 .