Question

the equivalent resistance of circuit is \frac{35}{3}.what is the value if resistor R​

Answer 1

$\large{\underline{\textsf{\textbf{ Answer : }}}}$

• R = $\sf { 20 }$ Ω

$\large{\underline{\textsf{\textbf{ Formulae \: to \: use : }}}}$

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

$\to \sf { \dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2}+ \dots \dots \dfrac{1}{R_n} }$

When resistors are connected in series combination, then equivalent resistance is given by :

$\to \sf { R_s = R_1 + R_2 + \dots \dots R_n }$

$\large{\underline{\textsf{\textbf{ Given \: Information }}}}$

• R₁ = 5 Ω
• R₂ = 10 Ω
• Equivalent resistance = $\sf { \dfrac{35}{3} }$ Ω

$\large{\underline{\textsf{\textbf{ To \: calculate }}}}$

• R [Resistor 3] = ?

$\large{\underline{\textsf{\textbf{ Steps : }}}}$

• R₂ and R are connected in parallel combination.
• Let Rₚ be the resultant resistance of R₂ and R.

Resultant resistance of R₂ and R will be given by :

$\to \sf { \dfrac{1}{R_p} = \dfrac{1}{R_2} + \dfrac{1}{R} }$

$\to \sf { \dfrac{1}{R_p} = \dfrac{1}{10} + \dfrac{1}{R} }$

$\to \sf { \dfrac{1}{R_p} = \dfrac{R+10}{10R} \: \Omega }$

$\to \sf { R_p = \dfrac{10R}{R+10} \: \Omega \dots \dots \rm {Equation \; (i) } }$

Resultant resistance of R₂ and R is $\sf { \dfrac{10R}{R+10} \: \Omega }$.

So, two resistances of 10 Ω and R connected in parallel are equal to a single resistance of $\sf { \dfrac{ 10R}{R+10} \: \Omega }$. Now, we can replace the two resistances R₂ and R by a single resistance of $\sf { \dfrac{10R}{R+10} \: \Omega }$.

The two resistors (R₁ and Rₚ) will be in series combination.

So,

→ Equivalent Resistance = Rₚ + R₁

Substituting the value of Rₚ from the equation (1).

→ Equivalent Resistance = $\sf { \Bigg \lgroup \dfrac{10R+ 5R + 50}{R+10} + 5 \Bigg \rgroup \Omega}$

→ Equivalent Resistance = $\sf { \Bigg \lgroup \dfrac{15R + 50}{R+10} \Bigg \rgroup \Omega}$

→ Equivalent Resistance = $\sf { \Bigg \lgroup \dfrac{15R + 50}{R+10} \Bigg \rgroup \Omega}$

But according to the question,

• Equivalent resistance is $\sf \dfrac{35}{3} \: \Omega$

So,

→ $\sf { \dfrac{35}{3} = \dfrac{15R+50}{R+10} }$

By cross multiplication,

→ 35(R+10) = 3(15R + 50)

→ 35R + 350 = 45R + 150

→ 350 - 150 = 45R - 35R

→ 200 = 10R

→ $\sf \dfrac{200}{10} \Omega$ = R

$\longrightarrow \boxed { \pmb {\sf {\red { R = 20 \: \Omega }}}}$

Therefore, the value of resistor R is $\sf { 20 }$ Ω.

_____________________________