Question

f the equivalent resistance of two resistors 6 Ω and X Ω connected in parallel with each other is 2 Ω then find the value of X.​

Answer 1

Answer:

3 Ω

Explanation:

As per the provided information in the given question, we have :

• R₁ = 6 Ω
• R₂ = X Ω
• Rₚ = 2 Ω (Equivalent resistance in parallel combination)

We are asked to calculate the value of X.

Basically, here we need an equation in order to solve for X. The formula to calculate the equivalent resistance when resistors are connected in parallel combination will act as a linear equation here. As it is know to us that,

When resistors are connected in parallel combination, the equivalent resistance is given by,

$\\ \longrightarrow \quad \boxed{\pmb{\sf { \dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dots \dfrac{1}{R_n} }} }$

Here,

$\longrightarrow \sf{\quad { \dfrac{1}{R_P} = \dfrac{1}{R_1} + \dfrac{1}{R_2} }}$

Substitute the values we have, in the formula.

$\longrightarrow \sf{\quad { \dfrac{1}{2} = \dfrac{1}{6} + \dfrac{1}{X} }}$

Taking the LCM in RHS and simplifying further.

$\longrightarrow \sf{\quad { \dfrac{1}{2} = \dfrac{X + 6}{6X} }}$

Now, using cross multiplication method to balance the equation.

$\longrightarrow \sf{\quad { 1(6X) = 2(X + 6) }}$

Performing multiplication on both sides.

$\longrightarrow \sf{\quad { 6X = 2X + 12 }}$

Transposing 2X from RHS to LHS.

$\longrightarrow \sf{\quad { 6X - 2X = 12 }}$

Performing subtraction in LHS.

$\longrightarrow \sf{\quad { 4X = 12 }}$

Transposing 4 from LHS to RHS. Its arithmetic operator will get changed.

$\longrightarrow \sf{\quad { X = \cancel{\dfrac{12}{4}} }}$

Dividing 12 by 4.

$\longrightarrow \quad \underline{\boxed { \textbf{\textsf{ X = 3 }}}}$

∴ The value of X is 3 Ω.