Question

What is the equivalent resistance across the points A and B in the given circuit.
​

Answer 1

Observe attachment for entire scenario , ( After have been modified )

10 Ω , 2.5 Ω are connected in parallel ,

Let it be R₁ ,

$:\implies \sf \dfrac{1}{R_1}=\dfrac{1}{10}+\dfrac{1}{2.5}$

$:\implies \sf \dfrac{1}{R_1}=\dfrac{5}{10}$

$:\implies \sf R_1=2\ \Omega\ \; \orange{\bigstar}$

Now ,

R₁ , 10 Ω are connected in series ,

Let it be R₂ ,

$:\implies \sf R_2 = R_1 +10\ \Omega$

$:\implies \sf R_2=2 +10$

$:\implies \sf R_{2}=12\ \Omega\ \; \green{\bigstar}$

Now ,

R₂ , 12 Ω are connected in parallel ,

Let it be R₃ ,

$:\implies \sf \dfrac{1}{R_3}=\dfrac{1}{R_2}+\dfrac{1}{12}$

$:\implies \sf \dfrac{1}{R_3}=\dfrac{1}{12}+\dfrac{1}{12}$

$:\implies \sf R_3 = 6\ \Omega\ \; \red{\bigstar}$

Now ,

R₃ , 10 Ω are connected in series ,

Let it be R₄ ,

$:\implies \sf R_4 =R_3 + 10$

$:\implies \sf R_4 =6+10$

$:\implies \sf R_4=16\ \Omega\ \; \blue{\bigstar}$

R₄ , 16 Ω are connected in parallel ,

Let it be R₅ ,

$:\implies \sf \dfrac{1}{R_{5}}=\dfrac{1}{R_4} + \dfrac{1}{16}$

$:\implies \sf \dfrac{1}{R_5}=\dfrac{1}{16}+\dfrac{1}{16}$

$:\implies \sf R_5=8\ \Omega\ \; \purple{\bigstar}$

So , Equivalent resistance b/w A and B in the given circuit is 8 Ω .