Question

The total resistance in the circuit when two resistors of resisitance 4 ohm each are connected in series and then parallel A) 8ohm ,1/8 ohm B) 8ohm, 2 ohm C) 16 ohm ,0.5 ohm D) 4 ohm ,4 ohm

Answer 1

Given ,

Resistance in series = R1 + R2 = 25 Ω

Resistance in parallel = 1/ R1 +1/ R2 = 1/4 Ω

= (R1 × R2)/R1 + R2 = 4 Ω

Suppose R1 = x Ω

Then R2 = (25 – x) Ω

(R1 × R2)/R1 + R2 = 4 Ω

(x) × (25 – x) / (x) + (25 – x) = 4 Ω

25x - x2 /25 = 4 Ω

- x2 + 25x = 100

(x^2 - 25x + 100) = 0

(x - 20x - 5x + 100) = 0

(x – 20) (x- 5) = 0

So x = 20 ohms or 5 Ω

IF R1 is taken as 20 Ω,

Then, R2 = 5 Ω

Answer : These two resisters are 20 Ω & 5 Ω

Answer 2

Answer:

In series:-

R = R1+R2

R = 4ohm + 4ohm.

R= 8ohm.

In parallel:-

I/R = 1/R1 + 1/R2

1/R = 1/4 + 1/4

1/R = 2/4 = 1/2ohm = 0.5ohm

Explanation:

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