Question

The equivalent resistance is connected in parallel is 4ohm and when it is connected in series is 25 ohm calculate the individual resistance​

Answer 1

Explanation:

A to Q

1/Req = 1/R1+1/R2

R1R2/R1+R2 = 4

R1R2 = 4R1+4R2 (1)

Req = R1+R2 = 25 (2)

R2 = 25-R1 (3)

From (1) and (2)

R1R2/4 = 25

R1R2 = 100 (4)

Putting Eq 3 in Eq 4

R1(25-R1) = 100

R1²-25R1+100 = 0

R1²-(20+5)R1+100 = 0

R1²-20R1-5R1+100 = 0

R1(R1-20)-5(R1-20) = 0

(R1-20)(R1-5) = 0

so, R1=20,5 and R2=5,20

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Answer 2

AnswEr :

Correct question :

Given that equivalent resistance of two resistors is 0.25Ω and equivalent resistance in series is 25Ω.

We have to find the individual resistance of two resistors.

Let the two resistors be R₁ and R₂

We know these two formulas :-

⇒ R (series) = R₁ + R₂ + ..............

⇒ R (parallel) = 1/R₁ + 1/R₂ + ...........

Substituting values,

⇒ 0.25 = 1/R₁ + 1/R₂

⇒ (R₂ + R₁)/R₁R₂ = 0.25

⇒ R₁ + R₂ = 0.25R₁R₂ ...(i)

Again substituting values,

⇒ R₁ + R₂ = 25 ...(ii)

Comparing both equations we get,

⇒ 0.25R₁R₂ = 25

⇒ R₁R₂ = 25/0.25

⇒ R₁R₂ = 100

⇒ R₁ = 100/R₂ ...(iii)

Now putting this value in (ii) :-

⇒ (100/R₂) + R₂ = 25

⇒ (100 + R₂²)/R₂ = 25

⇒ R₂² + 100 = 25R₂

⇒ R₂² - 25R₂ + 100 = 0

⇒ R₂² - 5R₂ - 20R₂ + 100 = 0

⇒ R₂ (R₂ - 5) - 20(R₂ - 5) = 0

⇒ (R₂ - 20)(R₂ - 5) = 0

⇒ R₂ = 20Ω or, R₂ = 5Ω

Now putting this value in (iii),

⇒ R₁ = 100/20 or, R₁ = 100/5

⇒ R₁ = 5Ω or, R₁ = 20Ω

∴ Individual resistance of two resistors are 5Ω and 20Ω