Question

please answer this question​

Answer 1

hope this answer will help you

Answer 2

Given :-

• Each resistor is of 3Ω

To find :-

• Equivalent resistance of circuit

Formula used :-

• $\sf Resistance\: in \:series=R_{(Series)}=R_1+R_2+R_3...$

• $\sf Resistance\:in\: parallel=\dfrac 1{R_{(Parallel)}}=\dfrac 1 R_1+\dfrac 1 R_2+\dfrac 1 R_3...$

Solution :-

In the given diagram, we can see that R₃ and R₂ are in series connection, their equivalent resistance will be::

>> Rₙ=R₂+R₃

>> Rₙ=3Ω+3Ω

>> Rₙ=6Ω

Now the connection or Rₙ (R₂ & R₃) and R₄ are in parallel connection, their equivalent resistance will be::

$\sf \mapsto \dfrac 1 R_x=\dfrac 1 R_n+\dfrac 1 R_4$

$\sf \mapsto \dfrac 1 R_x=\dfrac 1 {6\Omega}+\dfrac 1 {3\Omega}$

$\sf \mapsto \dfrac 1 R_x=\dfrac {1+2} {6\Omega}$

$\sf \mapsto \dfrac 1 R_x=\dfrac {3} {6\Omega}$

$\sf \mapsto \dfrac 1 R_x=\dfrac {\not 3} {\not 6\Omega}$

$\sf \mapsto \dfrac 1 R_x=\dfrac {1} {2\Omega}$

$\red{\sf \mapsto R_x=2\Omega}$

Now the connection of Rₓ (Rₙ & R₄), R₁ and R₅ are in series connection, their equivalent resistance will be::

⇒ Rₐ = Rₓ + R₁ + R₅

⇒ Rₐ = 2Ω + 3Ω + 3Ω

⇒ Rₐ=8Ω

So the equivalent resistance of the circuit is 8Ω.

[Omega will change into Ω when viewed from Web.]