Question

how can three resistors of resistance 2 ohm 3 ohm and 6 ohm be connected to give a total resistance of 4 ohm and 1 ohm

Answer 1

Answer:

Explanation:

Given :-

R₁ = 2 Ω      

R₂ = 3 Ω          

R₃ = 6 Ω      

Solution :-

(a)  When R2 and R3 are connected in parallel with R₁ in series we get                

⇒ 1/R = 1/R₂ + 1/R₃              

⇒ 1/R = 1/3 + 1/6                

⇒ 1/R = 1/2                

⇒ R = 2 Ω            

⇒ Resistance in series = R + R₁                

⇒ Resistance in series = 2 + 2 = 4 Ω

Hence, the total resistance of the circuit is 4 Ω.      

(b)  When R₁, R₂, R₃ are connected in parallel we get                

⇒ 1/R = 1/R₁ + 1/R₂ + 1/R₃                

⇒ 1/R = 1/2 + 1/3 + 1/6                

⇒ 1/R = 1 Ω.

Hence, the net equivalent resistance of the circuit is 1 Ω.

Answer 2

Answer:

There are three resistors of resistances 2 Ω, 3 Ω, and 6 Ω respectively.

(a) The following circuit diagram shows the connection of the three resistors.

Here, 6 Ω and 3 Ω resistors are connected in parallel.

Therefore, their equivalent resistance will be given by

This equivalent resistor of resistance 2 Ω is connected to a 2 Ω resistor in series.

Therefore, the equivalent resistance of the circuit = 2 Ω + 2 Ω = 4 Ω

Hence the total resistance of the circuit is 4 Ω.

(b) The following circuit diagram shows the connection of the three resistors.

All the resistors are connected in series. Therefore, their equivalent resistance will be given as

Therefore, the total resistance of the circuit is 1 Ω.