Question

What is the (a) highest (b) lowest total resistance, that can be secured by combinations of four coils of resistances 2 ohm, 4 ohm, 5 ohm and 10 ohm?​

Answer 1

Highest resistance is obtained if all the resistors are in series

Rs=2+4+5+10=21ohm

Lowest resistance is obtained when all are.in parallel

1/Rp=1/2+1/4+1/5+1/10

1/Rp=10+5+4+2/20

1/Rp=21/20

Rp=20/21=0.95 ohm

I hope this helps ( ╹▽╹ )

Answer 2

Answer:

Given :-                

R₁ = 4 Ω            

R₂ = 8 Ω                

R₃ = 12 Ω              

R₄ = 24 Ω          

Solution :-

(a) The highest resistance is when the resistances are connected in series.

Total resistance in series = R₁ + R₂ + R₃ + R₄                

Total resistance in series = 4 + 8 + 12 + 24                

Total resistance in series = 48 Ω          

Hence, the highest resistance is 48 Ω.          

(b)  The lowest resistance is when the resistances are connected in parallel        

1/R = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄          

1/R = 1/4 + 1/8 + 1/12 + 1/24          

1/R = 1/2 Ω

1/R = 1/2 Ω

R = 2 Ω          

Hence, the lowest resistance is 2 Ω  ..