Question

What is the equivalent resistance of the following
network of resistances?


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Answer 1

To find:-

equivalent resistance of series

solution:-

Starting from A first resistance is of 1Ω than moving further current is divided into two branches one branch Containing 3 resistors and other containing 1 resistor. hence those three are in series and to them 1 is in parallel.

First solving the two branches.

$\mathtt{ \frac{1}{r} = \frac{1}{r1} + \frac{1}{r2} }$

r1 = 1Ω+1Ω+1Ω=3Ω

r2=1Ω

putting the values

$\mathtt{ \frac{1}{r} = \frac{1}{3} + \frac{1}{1} }$

hence r= 3/4Ω

Now r and other two single resistors are in series hence total resistance will be submition of all

$\large\mathtt{R_{eq} = 1 Ω+ \frac{3}{4}Ω + 1Ω }$

$\large\mathtt{R_{eq} = \frac{11}{4} }$

$\LARGE\boxed{ \mathtt{Answer = \frac{11}{4}Ω }}$