Question

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Answer 1

Answer :

We have to find equivalent resistance between A and C$.$

★ Equivalent resistance of the series connection is given by

• R = R₁ + R₂ + ... + Rₙ

★ Equivalent resistance of the parallel connection is given by

• 1/R = 1/R₁ + 1/R₂ + ... + 1/Rₙ

Let's solve it step by step!

First see the upper part, two resistors of 4Ω and 12Ω are connected in series.

Hence equivalent of that branch will be

• 4 + 12 = 16Ω

After solving this, 16Ω and 2Ω come in parallel. Hence equivalent of this part will be

• (16 × 2) / (16 + 2)
• 32/18
• 1.77Ω

Finally three resistors of 5Ω, 1.77Ω and 4Ω come into series.

Hence net equivalent resistance will be

→ R = 5 + 1.77 + 4

→ R = 10.77Ω

Answer 2

Answer:

We have to find equivalent resistance between A and C:

★ Equivalent resistance of the series connection is given by

• R = R₁ + R₂ + ... + Rₙ

★ Equivalent resistance of the parallel connection is given by

• 1/R = 1/R₁ + 1/R₂ + ... + 1/Rₙ

step by step solution!

First see the upper part, two resistors of 4Ω and 12Ω are connected in series.

Hence equivalent of that branch will be

4 + 12 = 16Ω

After solving this, 16Ω and 2Ω come in parallel. Hence equivalent of this part will be

(16 × 2) / (16 + 2)

32/18

1.77Ω

Finally three resistors of 5Ω, 1.77Ω and 4Ω come into series.

Hence net equivalent resistance will be

→ R = 5 + 1.77 + 4

→ R = 10.77Ω