Parallel Resistors

When $n$ resistors are connected in parallel, the equivalent resistance \(R_{eq}is given by:

\[\frac{1}{R_{eq}}=\sum _{ i=1 }^{ n }{ \frac { 1 }{ { R }_{ i } } } \]

For n resistors of equal resistance $R$ , we have:

$\frac{1}{R_{eq}}=\frac{n}{R} \Rightarrow \boxed{R_{eq}=\cfrac{R}{n}}$

As it was given equal resistance let us assume each resistance to be R. Now let us consider only 2 resistors in parallel we get (R.R)/(R+R)= R/2 As we did for 2 resistors denominator was 2, if we do for n no: of resistors we get R/n

Effective resistance:

1/Reff = 1/R1 + 1/R2+......+ 1/Rn

since equal resistance R1=R2=........=Rn

1/Reff = 1/R+1/R+......+1/R

1/Reff = n/R

or Reff = R/n

Make it like this:

The formula is:

$\frac{1}{R1}$ + $\frac{1}{R2}$ ... + $\frac{1}{R(N-1)}$ $\frac{1}{RN}$

Because of R1=R2=R(N-1)=RN so we make all R(1,2,...,N} to be R:

$\frac{1}{R}$ + $\frac{1}{R}$ + $\frac{1}{R}$ + $\frac{1}{R}$ + ..... = $\frac{n}{R}$

n= $\sum_{1}$

Then do the magic:

$\frac{1}{\frac{n}{R}}$ = $\frac{R}{n}$