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-1
Judge the equivalent resistance when the following are connected in
parallel-
(a) 1 ohm and 106 ohm,
(b) 1 ohm and 103 ohm, and 106 ohm.
(a) The net resistance in parallel is given by
1/𝑅=1/𝑅₁ +1/𝑅₂
Here, 𝑅₁=1 Ω and 𝑅₂=10⁶ Ω, So,
1/𝑅 =1/1 + 1/10⁶ = 10⁶+1/10⁶ ⟹ 𝑅=10⁶ /10⁶ +1≈1Ω
(b) The net resistance in parallel is given by
1/𝑅=1/𝑅₁ +1/𝑅₂ +1/𝑅₃
Here, 𝑅₁ =1 Ω, 𝑅₂=10⁶ Ω and 𝑅3=10⁶ Ω, So,
1/𝑅 = 1/1+1/10³+1/10⁶ = 10⁶ +10³+1/10⁶ =1001001/1000000 ⟹𝑅=1000000/1001001= 0.999Ω ≈ 1Ω
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When the resistances are connected in parallel, the equivalent resistance is smaller than the smallest individual resistance.
(i) Equivalent resistance < 1 Ω.
(ii) Equivalent resistance < 1 Ω.
Since 1/R=1/R1+1/R2+1/R3+..+1/Rn
when resistors are connected in parallel
(a) 1 Ω and 106 Ω
Thus, 1/R=1/1Ω+1/106Ω=106+1/106Ω=107/106Ω
Thus, R=106/107Ω=0.99Ω
Thus, equivalent resistance of 1Ω and 106Ω are connected in parallel = 0.99Ω
(b) 1 Ω and 103 Ω, and 106 Ω
Thus, 1/R=1/1Ω+1/103Ω+1/106Ω=(10918+106+103)/103×106Ω=11127/10918Ω
Thus, R=10918/11127Ω=1.02Ω
Thus, equivalent resistance of 1 Ω, 103 Ω and 106 Ω are connected in parallel = 1.02Ω