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.

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.

(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

For more answers visit to website:

https://www.tiwariacademy.com/ncert-solutions/class-10/science/chapter-12/

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 Ω1 Ω, 103 Ω and 106 Ω are connected in parallel = 1.02Ω

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