Question

The effective resistances are 10Ω, 2.4Ω when two resistors are connected in series and parallel. What are the resistances of individual resistors?

Answer 1

Hello there!
Let's start with assuming two resistances R1 and R2 respectively.

R1 + R2 = 10

As when two resistances are in series the effective resistance is the sum of the two resistances.

1/R1 + 1/R2 = 1/2.4

As when two resistances are in parallel the effective resistance is calculated as given above.

Now we rearrange the above equation.

(R1 × R2)/(R1 + R2) = 2.4

(R1 × R2) = 2.4 × (R1 + R2)

(R1 × R2) = 2.4 × 10 (From the first equation)

(R1 × R2) = 24

(R1 + R2) = 10 (From the first equation)

R1 + 24/R1 = 10

(R1)^2 - 10×R1 +24 = 0

(R1 - 4)(R1 - 6) = 0

Thus R1 can be of 4 or 6 ohms. Thus if R1 is of 4 ohms then R2 is of 6 ohms and vice versa.

Thus the two resistances are of 4 and 6 ohms.

Answer 2

Answer:

Concept:

Effective resistance when connected in series and parallel.

Explanation:

Given:

$\mathrm{R}_{\mathrm{S}}=10 \Omega ; \mathrm{R}_{\mathrm{P}}=2.4 \Omega$

Find:

$\mathrm{R}_{1}=? \\\mathrm{R}_{2}=?$

Solution:

Effective resistance when connected in series and parallel.

$\mathrm{R}_{\mathrm{S}}=\mathrm{R}_{1}+\mathrm{R}_{2}=10$   equation(1)

$\frac{1}{R_{P}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=\frac{R_{1}+R_{2}}{R_{1} R_{2}}$

$\therefore \mathrm{R}_{\mathrm{P}}=\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{\mathrm{R}_{1}+\mathrm{R}_{2}}=2.4$   equation(2)

From equations (1) and (2)

$\begin{aligned}&\frac{\mathrm{R}_{1} \mathrm{R}_{2}}{10}=2.4 \\&\Rightarrow \mathrm{R}_{1} \mathrm{R}_{2}=24 \\&\therefore \mathrm{R}_{2}=\frac{24}{\mathrm{R}_{1}}\end{aligned}$

$Substituting \mathrm{R}_{2} in eqn (1) \begin{aligned}&R_{1}+\frac{24}{R_{1}}=10 ; R_{1}^{2}+24=10 R_{1} \\&R_{1}^{2}-10 R_{1}+24=0 \\&\left(R_{1}-6\right)\left(R_{1}-4\right)=0 \\&\therefore R_{1}=6 \Omega \text { or } R_{1}=4 \Omega \\&R_{2}=\frac{24}{6} \text { or } R_{2}=\frac{24}{4} \\&=4 \Omega . \text { or }=6 \Omega \\&R_{1}=6 \Omega, R_{2}=4 \Omega .\end{aligned}$

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