Tricky Solutions For Equivalent Series-parallel Resistance Problems With Examples

Tricky Solutions For Equivalent Series-parallel Resistance Problems With Examples

Examples Based On Part-1 & Part-2 :-

Ques.1) Find the equivalent resistance between A & B for the given network ?

 
For solving this problem, we follows same step as we discussed in Part-2.

So, R_AB=18.611Ω

Ques.2) Find the equivalent resistance between B & C for the given network.

                 

Hints:-

Explanation:- 

For finding R_bc,we firstly search the last box.Here,three boxes formed as shown in fig.For this circuit,box 1 & 2 both seemed last box from pt B & C. So,we can reduced both the boxes at same time or do it one-by-one,as you like.

For box 1,All are in series.

So,

R_series=R₁+R₂+R₃=40+20+10=70Ω

For box 3,

R_series= R₁+R₂=12+8=20Ω

Again,

both the boxes become last boxes.In box 1,70Ω & 30Ω are in parallel.

Then, R_parallel=R₁||R₂= 21Ω

Now,

In box 3,20Ω & 20Ω  are in parallel and also identical.

So,R_parallel becomes half.

i.e,10Ω.

After doing these all only one box remains,in which 50Ω & 10Ω are in series.

R_s=50+10=60Ω

At last 21Ω & 60Ω become parallel to each other.

So,R_bc=21||60=15.55Ω.

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