Question

There are 3 resistors of magnitude 12 ohm,15 ohm,20 ohm. find its resistance when connected in series and when connected in parallel

Answer 1

Answer :

Given -

• Three resistors are connected
• The magnitude of First resistor = 13 ohm
• The magnitude of second resistor = 15 ohm
• The magnitude of third resistor = 20 ohm

To Find -

• Net resistance in parallel connection ?
• Net resistance in series connection ?

Answer -

For calculating the net resistance in series connection : Rs = R1 + R2 + R3

⇒ 12 + 15 + 20

⇒ 47 ohm

For calculating the net resistance in parallel connection : 1/Rp = 1/R1 + 1/R2 + 1/R3

⇒ 1/Rp = 1/12 + 1/15 + 1/20

⇒ 1/Rp = 1/12 + (20 + 15)/300

⇒ 1/Rp = 1/12 + (35/300)

⇒ 1/Rp = (300 + 420)/3,600

⇒ 1/Rp = 720/3,600

⇒ 1/Rp = 0.2 ohm

⇒ Rp = 0.2/1

⇒ Rp = 5 ohm

So, the net resistance in series connection is 47 ohm whereas the net resistance in the parallel connection is is 5 ohm.

Answer 2

AnswEr :

Given that there are 3 resistors of magnitude 12Ω, 15Ω and 20Ω

We have to find equivalent resistance if connected in series and if connected in parallel.

Let's find equivalent resistance in series :

We know,

Rₛ = R₁ + R₂ + R₃ + ...............

⟼ Rₛ = 12 + 15 + 20

⟼ Rₛ = 47Ω

∴ Equivalent resistance in series = 47Ω

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Now let's find equivalent resistance in parallel :

We know that,

1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃ + .................

⟼ 1/Rₚ = 1/12 + 1/15 + 1/20

⟼ 1/Rₚ = (5 + 4 + 3)/60

⟼ 1/Rₚ = 12/60

⟼ 1/Rₚ = 1/5

⟼ Rₚ = 5Ω

∴ Equivalent resistance in parallel = 5Ω