Question

Find the equivalent resistance between A and B !

Need correct explanation, others would be reported !​​

Answer 1

Answer:

The equivalent resistance between A and B is 9Ω.

Explanation:

To know :

• When resistors of resistances R₁ , R₂ , R₃ .... are connected in series, the equivalent resistance is given by

       R = R₁ + R₂ + R₃ + ...

• When resistors of resistances R₁ , R₂ , R₃ .... are connected in parallel, the equivalent resistance is given by

       1/R = 1/R₁ + 1/R₂ + 1/R₃ + ...

Solution :

[Refer to the attachment]

First, let's find the equivalent resistance between A and P :

• The upper two resistors are connected in series.

 Equivalent resistance of the two resistors = 3Ω + 3Ω = 6Ω

• The lower two resistors are connected in series.

 Equivalent resistance of the two resistors = 3Ω + 3Ω = 6Ω

• These two combinations are connected in parallel.

Let the equivalent resistance between A and P be R[AP]

   $\sf \dfrac{1}{R[AP]} =\dfrac{1}{6}+\dfrac{1}{6} \\\\ \sf \dfrac{1}{R[AP]}=\dfrac{1+1}{6} \\\\ \sf \dfrac{1}{R[AP]}=\dfrac{2}{6} \\\\ \sf \dfrac{1}{R[AP]}=\dfrac{1}{3} \\\\ \sf \longrightarrow R[AP]=3 \Omega$

Find the equivalent resistance between P and Q :

• The upper two resistors are connected in series.

 Equivalent resistance of the two resistors = 3Ω + 3Ω = 6Ω

• The lower two resistors are connected in series.

 Equivalent resistance of the two resistors = 3Ω + 3Ω = 6Ω

• These two combinations are connected in parallel.

Let the equivalent resistance between P and Q be R[PQ]

   $\sf \dfrac{1}{R[PQ]} =\dfrac{1}{6}+\dfrac{1}{6} \\\\ \sf \dfrac{1}{R[PQ]}=\dfrac{1+1}{6} \\\\ \sf \dfrac{1}{R[PQ]}=\dfrac{2}{6} \\\\ \sf \dfrac{1}{R[PQ]}=\dfrac{1}{3} \\\\ \sf \longrightarrow R[PQ]=3 \Omega$

Find the equivalent resistance between Q and B :

• The upper two resistors are connected in series.

 Equivalent resistance of the two resistors = 3Ω + 3Ω = 6Ω

• The lower two resistors are connected in series.

 Equivalent resistance of the two resistors = 3Ω + 3Ω = 6Ω

• These two combinations are connected in parallel.

Let the equivalent resistance between A and P be R[QB]

   $\sf \dfrac{1}{R[QB]} =\dfrac{1}{6}+\dfrac{1}{6} \\\\ \sf \dfrac{1}{R[QB]}=\dfrac{1+1}{6} \\\\ \sf \dfrac{1}{R[QB]}=\dfrac{2}{6} \\\\ \sf \dfrac{1}{R[QB]}=\dfrac{1}{3} \\\\ \sf \longrightarrow R[QB]=3 \Omega$

The equivalent resistance between A and B :

R[AP] , R[PQ] and R[QB] are connected in series.

Hence,

The equivalent resistance between A and B = R[AP] + R[PQ] + R[QB]

   = 3Ω + 3Ω + 3Ω

   = 9Ω