Question

Three resistors of 6 Ω, 3 Ω and 2 Ω are connected together so that the total resistance is greater than 6 Ω but less than 8 Ω. Draw a diagram to show this arrangement and calculate its total resistance.

SOLVE WITH WORKING
Please solve fast
NO SPAM​

Answer 1

Answer :

• The total resistance in the whole circuit is 7.2 Ω.

Explanation :

Given :

• Resistance in the resistor, R₁ = 6 Ω
• Resistance in the resistor, R₂ = 3 Ω
• Resistance in the resistor, R₃ = 2 Ω
• Resistance in the circuit is greater than 6 Ω and lesser than 8 Ω.

To find :

• Total resistance in the circuit, Rₙ = ?

Knowledge required :

• Formula for Total resistance in case of a series circuit :

⠀⠀⠀⠀⠀⠀⠀⠀⠀Rₑ = R₁ + R₂ + R₃ + ... + Rₙ⠀

Where,

• Rₑ = Total resistance in the series circuit.
• R = Resistance in the given circuits.

• Formula for Total resistance in case of a pa5al5 circuit :

⠀⠀⠀⠀⠀⠀⠀⠀⠀1/Rₑ = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/Rₙ⠀

Where,

• Rₑ = Total resistance in the parallel circuit.
• R = Resistance in the given circuits.

Assumption :

Let us assume that the resistor R₂ and R₃ are connected in parallel circuit and the resistor R₁ is connected in series circuit.

Solution :

First let us find the total resistance in the series circuit.

By using the formula for total resistance in a parallel circuit and substituting the values in it, we get :

⠀⠀=> 1/Rₑ = 1/R₁ + 1/R₂ + 1/R₃ + ... + 1/Rₙ

⠀⠀=> 1/Rₑ = 1/R₁ + 1/R₂

⠀⠀=> 1/Rₑ = 1/3 + 1/2

⠀⠀=> 1/Rₑ = (2 + 3)/6

⠀⠀=> 1/Rₑ = 5/6

⠀⠀=> Rₑ = 6/5

⠀⠀=> Rₑ = 1.2

⠀⠀⠀⠀⠀⠀⠀∴ Rₑ = 1.2 Ω

Hence the total resistance in the parallel circuit is 1.2 Ω.

Now fet's find out the total resistance in the series circuit :

⠀⠀=> Rₑ = R₁ + R₂ + R₃ + ... + Rₙ

⠀⠀=> Rₑ = R₁ + R₂

⠀⠀=> Rₑ = 6 + 1.2

⠀⠀=> Rₑ = 7.2

⠀⠀⠀⠀⠀⠀⠀∴ Rₑ = 7.2 Ω

Hence the total resistance in the circuit and is 7.2 Ω.