Question

HoW CaN ThRee ReSiStOrS Of ReSiStAnCeS 2 OhM , 3 OhM AnD 6 OhM Be CoNnEcTeD To GiVe A ToTaL ReSiStAnCe Of (a) 4 OhM , (b) 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

a) 3ohm and 6ohm connect in parallel

1/Rp=1/3+1/6=2+1/6

Rp =6/3=2

then in series

2+2=4ohm

b) 2,3and 6ohm connect into parallel

1/Rp=1/2+1/3+1/6

1/Rp = 3+2+1/6

Rp =6/6=1ohm

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