Question

equivalent resistance between the points A and B in given figure please answer me by step by step method ​

Answer 1

Answer:

2 Ω

Explanation:

As it can be seen in the figure, resistances AB and BC are connected in series.

For the top part:

So, the effective resistance = Sum of resistance in series.

→ Effective resistance = 3 Ω + 3 Ω

→ Effective resistance = 6 Ω

For below part:

Similarly, AD and CD are connected in series.

Effective resistance = Sum of resistance in series

→ Effective resistance = 3 Ω + 3 Ω

→ Effective resistance = 6 Ω

Now,

These 2 effective resistances are connected in parallel with resistance AC.

So, let us consider

• R₁ = 6 Ω
• R₂ = 6 Ω
• R₃ = 6 Ω

The effective resistance in parallel combination would be:

$\tt{\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} }$

$\implies \sf{ \dfrac{1}{R} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} }$

$\implies \sf{ \dfrac{1}{R} = \dfrac{3}{6} }$

$\implies \sf{ \dfrac{1}{R} = \dfrac{1}{2} }$

$\implies \boxed{ \bf{R = 2 \Omega} }$

Therefore, the total equivalent resistance is 2 Ω

Answer 2

Answer:

Answer:

2 Ω

\begin{gathered}\\\end{gathered}

Explanation:

As it can be seen in the figure, resistances AB and BC are connected in series.

\begin{gathered}\\\end{gathered}

For the top part:

So, the effective resistance = Sum of resistance in series.

→ Effective resistance = 3 Ω + 3 Ω

→ Effective resistance = 6 Ω

\begin{gathered}\\\end{gathered}

For below part:

Similarly, AD and CD are connected in series.

Effective resistance = Sum of resistance in series

→ Effective resistance = 3 Ω + 3 Ω

→ Effective resistance = 6 Ω

\begin{gathered}\\\end{gathered}

Now,

These 2 effective resistances are connected in parallel with resistance AC.

So, let us consider

R₁ = 6 Ω

R₂ = 6 Ω

R₃ = 6 Ω

\begin{gathered}\\\end{gathered}

The effective resistance in parallel combination would be:

\begin{gathered}\tt{\dfrac{1}{R} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} } \\ \\ \end{gathered}

R

1

=

R

1

1

+

R

2

1

+

R

3

1

\begin{gathered} \implies \sf{ \dfrac{1}{R} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} } \\ \\ \end{gathered}

⟹

R

1

=

6

1

+

6

1

+

6

1

\begin{gathered} \implies \sf{ \dfrac{1}{R} = \dfrac{3}{6} } \\ \\ \end{gathered}

⟹

R

1

=

6

3

\begin{gathered}\implies \sf{ \dfrac{1}{R} = \dfrac{1}{2} } \\ \\ \end{gathered}

⟹

R

1

=

2

1

\begin{gathered} \implies \boxed{ \bf{R = 2 \text{\O}mega} } \\ \\ \end{gathered}

⟹

R=2Ømega