Question

As shown in figure,the resistance are connected with 12V battery.
Determine - (a) equivalent curcit resistance
(b) current flowing through the curcit ​

Answer 1

Answer :

• The equivalent resistance of the circuit = 18 Ω

• The current flowing through the circuit = 2/3 A

Explanation :

Given :

• R₁ = 10 Ω
• R₂ = 40 Ω
• R₃ = 30 Ω
• R₄ = 20 Ω
• R₅ = 60 Ω
• V = 12 V

To find :

• the equivalent circuit resistance

• the current flowing through the circuit

Solution :

 

To know :

• When resistors R₁ , R₂ , .... Rₙ are connected in parallel, the equivalent resistance is given as 1/Rₚ = 1/R₁ + 1/R₂ + .... + 1/Rₙ

• When resistors R₁ , R₂ , .... Rₙ are connected in series, the equivalent resistance is given as Rₛ = R₁ + R₂ + .... + Rₙ

In the given circuit,

• R₁ and R₂ are connected in parallel
• R₃ , R₄ and R₅ are connected in parallel

Let R' be the equivalent resistance of the resistors R₁ and R₂.

   $\sf \dfrac{1}{R'}=\dfrac{1}{R_1}+\dfrac{1}{R_2} \\\\ \sf \dfrac{1}{R'}=\dfrac{1}{10}+\dfrac{1}{40} \\\\ \sf \dfrac{1}{R'}=\bigg(\dfrac{1}{10} \times \dfrac{4}{4} \bigg)+\dfrac{1}{40} \\\\ \sf \dfrac{1}{R'}=\dfrac{4}{40}+\dfrac{1}{40} \\\\ \sf \dfrac{1}{R'}=\dfrac{4+1}{40} \\\\ \sf \dfrac{1}{R'}=\dfrac{5}{40} \\\\ \sf \dfrac{1}{R'}=\dfrac{1}{8} \\\\ \sf R'=8 \Omega$

Let R'' be the equivalent resistance of the resistors R₃ , R₄ and R₅

  $\sf \dfrac{1}{R''}=\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5} \\\\ \sf \dfrac{1}{R''}=\dfrac{1}{30}+\dfrac{1}{20}+\dfrac{1}{60} \\\\ \sf \dfrac{1}{R''}=\bigg(\dfrac{1}{30} \times \dfrac{2}{2} \bigg)+\bigg(\dfrac{1}{20} \times \dfrac{3}{3}\bigg)+\dfrac{1}{60} \\\\ \sf \dfrac{1}{R''}=\dfrac{2}{60}+\dfrac{3}{60}+\dfrac{1}{60} \\\\ \sf \dfrac{1}{R''}=\dfrac{2+3+1}{60} \\\\ \sf \dfrac{1}{R''}=\dfrac{6}{60} \\\\ \sf \dfrac{1}{R''}=\dfrac{1}{10} \\\\ \sf R''=10 \Omega$

Now, R' and R'' are in series combination.

 Let the equivalent resistance of the given circuit be R

R = R' + R''

R = 8 Ω + 10 Ω

R = 18 Ω

By using Ohm's law,

V = IR

12 = I × 18

I = 12/18

I = 2/3 A

 

Therefore,        

The equivalent resistance of the circuit = 18 Ω

The current flowing through the circuit = 2/3 A