Question

.What is the (a) highest (b) lowest total resistance, that can be secured by combinations of four coils of resistances 4 ohm, 8 ohm, 12 ohm and 24 ohm?

Answer 1

Highest will be by series = 4+12+8+24=48 ohms

Lowest will be by parallel = 1/R =1/4+1/8+1/24+1/12 = 12/24 = 1/2 
                                         so R = 2

Answer 2

Answer:

Explanation:

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 = 12/24          

1/R = 1/2 Ω

R = 2 Ω          

Hence, the lowest resistance is 2 Ω.