NCERT Solution for Class 10 Science Chapter 12
Electricity
NCERT Books for Session 2022-2023
CBSE Board and UP Board
Intext Questions
Page No-216
Questions No-5
What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 ohm, 8 ohm, 12 ohm, 24 ohm?
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The total resistance of resistors or coils in series is the sum of their individual resistances, and for resistors or coils in parallel, the reciprocal of the total resistance is the sum of the reciprocals of their individual resistances.
(a) Highest Total Resistance:
For the highest total resistance, you would connect the resistors in series because the total resistance in a series connection is the sum of the individual resistances.
R_total, series = R₁ + R₂ + R₃ + R₄
R_total, series = 4 + 8 + 12 + 24 = 48Ω
So, the highest total resistance is 48Ω when the resistors are connected in series.
(b) Lowest Total Resistance:
For the lowest total resistance, you would connect the resistors in parallel because the total resistance in a parallel connection is given by the reciprocal of the sum of the reciprocals of the individual resistances.
1/R_total, paraller = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄
1/R_total, paraller = 6/24 + 3/24 + 2/24 + 1/24
1/R_total, parallel = 12/24
total, parallel = 2Ω
Rtotal, parallel = 2Ω
(a) When all the resistors are connected in parallel
Then the 1/R=1/4Ω+1/8Ω+1/12Ω+1/24Ω=(6+3+2+1)/24Ω=12/24Ω=1/2Ω
Thus, R = 2 Ω
(b) When all the resistors are connected in series
Thus, total resistance = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω
Thus, highest resistance = 48 Ω
Lowest resistance = 2 Ω
(i) Highest resistance can be obtained by connecting the four coils in series.
Then, R = 4Ω + 8Ω + 12Ω + 24Ω = 48Ω
(ii) Lowest resistance can be obtained by connecting the four coils in parallel.
1/r = 1/4 +1/8+1/12+1/24=12/24=1/2
R= 2ohm.
Connecting resistors in series always gives maximum resistance and parallel gives minimum resistance.
(a) The highest total resistance is given by
𝑅 =𝑅₁+𝑅₂+𝑅₃+𝑅₄
=4 Ω+ 8 Ω+ 12 Ω+ 24 Ω
=48 Ω
(b) The lowest total resistance is given by
1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃+1/𝑅₄
1𝑅=1/4+1/8+1/12+1/24
= 6+3+2+1/24 = 12/24
⟹𝑅=24/12=2 Ω