What is the (a) highest (b) lowest total resistance, that can be secured by combinations of four coils of resistances 2 ohm 10 ohm 14 ohm 6 ohm
Answers
Answer:
Ans: Given resistances R1 =4 Ω, R2=8 Ω, R3=12Ω, and R4=24Ω.
(a) If these coils are connected in series, then the equivalent resistance will
be the highest, as R=R1+R2+R3+R4
∴ R=4+8+12+24 = 24Ω
(b) If these coils are connected in parallel, then the equivalent resistance will be the lowest, given by
⇒R=2 Ω
∴ 2 Ω is the lowest total resistance
Given: Combination of four coil of resistances,
= 2Ω
= 10Ω
= 14Ω
= 6Ω
To Find: a) Highest Total Resistance
b) Lowest Total Resistance
Solution:
a)Highest Total Resistance :-
The highest total resistance can be find by connecting all the four resistances of coil in Series combination as,
= + + +
Now,
By applying above equation we get,
= + + +
= 2Ω + 10Ω + 14Ω + 6Ω
∴ = 32Ω
∴ The Highest Resistance by the combination of the 4 coils = 32Ω
b) Lowest Total Resistance :
The lowest total resistance can be find by connecting all the four resistances of coil in Parallel combination as,
= + + +
Now,
By applying above equation we get,
= + + +
= Ω + Ω + Ω + Ω
∴ ≈ 1.19 Ω
∴ Lowest Total Resistance by the combination of the 4 coils = 1.19 Ω
so, a) Highest Total Resistance :- 32 Ω
b) Lowest Total Resistance :- ≈ 1.19 Ω