Question

Find the total resistance.

Answer 1

Answer:

please follow the above attachment for the solution

Answer 2

To Find:-

• The total effective resistance in the diagram provided.

Note:-

• It's necessary to the label each resistor in a circuit. (As shown in the attachment)

Solution:-

From the attachment we have:-

• R₁ = 2Ω
• R₂ = 5Ω
• R₃ = 20Ω
• R₄ = 1Ω
• R₅ = 3Ω
• R₆ = 4Ω
• R₇ = 6Ω
• R₈ = 12Ω

Now,

Here we can see that:-

• Resistors R₂ and R₃ are parallel
• Resistors R₆, R₇ and R₈ are also parallel

Hence, Firstly let us find the effective resistance in the resistors connected in parallel combination.

We know,

Effective resistance in a circuit is represented as follows:-

• $\sf{\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + . . . . . . . . . . . . . . . \dfrac{1}{R_n}}$

For resistors R₂ and R₃:-

$\sf{\dfrac{1}{R_p} = \dfrac{1}{5} + \dfrac{1}{20}}$

=> $\sf{\dfrac{1}{R_p} = \dfrac{4+1}{20}}$

=> $\sf{\dfrac{1}{R_p} = \dfrac{5}{20}}$

=> $\sf{\dfrac{1}{R_p} = \dfrac{1}{4}}$

=> $\sf{R_p = 4\Omega}$

For resistors R₆, R₇ and R₈:-

= $\sf{\dfrac{1}{R_t} = \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{12}}$

=> $\sf{\dfrac{1}{R_t} = \dfrac{3+2+1}{12}}$

=> $\sf{\dfrac{1}{R_t} = \dfrac{6}{12}}$

=> $\sf{\dfrac{1}{R_t} = \dfrac{1}{2}}$

=> $\sf{R_t = 2\Omega}$

Now,

From the second attachment we can see that resistors R₁, R₅ and Rₜ are in series.

Also, resistors R₁ and Rₚ are in series.

We know,

Effective resistance in series combination is represented by:-

• Rₛ = R₁ + R₂ . . . . . . . . . . . . Rₙ

Hence,

For resistors R₄, R₅ and Rₜ:-

$\sf{R_a = R_4 + R_3 + R_t}$

= $\sf{R_a = 1 + 3 + 2}$

= $\sf{R_a = 6\Omega}$

For resistors R₁ and Rₚ:-

$\sf{R_b = R_1 + R_p}$

= $\sf{R_b = 2 + 4}$

= $\sf{R_b = 6\Omega}$

Finally,

The resistors $\sf{R_a}$ and $\sf{R_b}$ are parallel (See 3rd attachment).

Hence,

The effective resistance across the circuit is :-

$\sf{\dfrac{1}{R_{AB}} = \dfrac{1}{6} + \dfrac{1}{6}}$

=> $\sf{\dfrac{1}{R_{AB}} = \dfrac{1+1}{6}}$

=> $\sf{\dfrac{1}{R_{AB}} = \dfrac{2}{6}}$

=> $\sf{\dfrac{1}{R_{AB}} = \dfrac{1}{3}}$

=> $\sf{R_{AB} = 3 \Omega}$

∴ The effective resistance across the circuit is 3Ω.

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