Question

Two resistance when connected in parallel give resultant value of 4 ohm(Ω),when connected in series the value becomes 25 ohm(Ω). What will be value of each resistance​

Answer 1

Answer :-

The required value of the resistances are 20 Ω and 5 Ω .

Explanation :-

For parallel connection :-

→ Equivalent resistance = 4 Ω

⇒ 1/R = 1/R₁ + 1/R₂

⇒ 1/4 = 1/R₁ + 1/R₂

⇒ R₁R₂/R₁ + R₂ = 4 Ω ---(1)

For series connection :-

→ Equivalent resistance = 25 Ω

⇒ R₁ + R₂ = 25 Ω ---(2)

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Now from eq.1, we have :-

⇒ R₁R₂ = (R₁ + R₂)(4 Ω)

⇒ R₁R₂ = (25 Ω)(4 Ω)

⇒ R₁R₂ = 100 Ω²

Now, let's calculate value of 'R₁ - R₂' :-

(a - b)² = (a + b)² - 4ab

⇒ (R₁ - R₂)² = (R₁ + R₂)² - 4R₁R₂

⇒ (R₁ - R₂)² = (25)² - 4(100)

⇒ R₁ - R₂ = √225

⇒ R₁ - R₂ = 15 Ω ----(3)

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Adding eq.2 and eq.3, we get :-

⇒ R₁ + R₂ + R₁ - R₂ = (25 + 15)

⇒ 2R₁ = 40

⇒ R₁ = 40/2

⇒ R₁ = 20 Ω

Putting value of 'R₁' in eq.2 :-

⇒ 20 + R₂ = 25

⇒ R₂ = 25 - 20

⇒ R₂ = 5 Ω

Answer 2

Answer:

20 Ω and 5 Ω are the values of resistances.

Explanation:

Given :-

The resultant resistance when two resistances are connected in parallel = 4 Ω

The resultant resistance when two resistances are connected in series = 25 Ω

To find :-

Value of each resistance.

Solution :-

Resistance value when connected in parallel :

$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{yR_2}$

$\rightarrow\frac{1}{4}=\frac{1}{R_1}+\frac{1}{R_2}\;(1)$

Resistance value when connected in series :

$R_s=R_1+R_2$

$\rightarrow25=R_1+R_2$

$\rightarrow R_2=25-R_1$ $(2)$

Substitute $(2)$ in $(1)$ :

$\frac{1}{4}=\frac{1}{R_1}+\frac{1}{25-R_1}$

$\frac{1}{4}= \frac{(25-R_1)+R_1}{R_1(25-R_1)}$

$25R_1-R_1^2=100-4R_1+4R_1$

$R_1^2-25R_1=100$

$R_1^2-20R_1-5R_1+100=0$

$(R_1-20)(R_1-5)=0$

∴ $R_1=20,5$

∴,

When $R_1$ is 25 Ω, $R_2$ is 5 Ω.

When $R_1$ is 5 Ω, $R_2$ is 25 Ω.