Question

how can three resistances of 2 ohm 3 ohm and 6 ohm be connected to give the total resistance of 4 ohm and 1 Ohm​

Answer 1

Explanation:

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

R

Answer 2

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.