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-2
An electric lamp of 100 ohm, a toaster of resistance 50 ohm, and a water filter of resistance 500 ohm are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
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Given that the electric lamp of 100 Ω, a toaster of resistance 50 Ω and water filter of resistance 500 Ω are connected in parallel.
The net resistance in parallel is given by
1/𝑅 = 1/𝑅₁+1/𝑅₂+1/𝑅₃
Here, 𝑅₁=100 Ω, 𝑅₂=50 Ω and 𝑅₃=500 Ω, So,
1/𝑅 = 1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.25 Ω
Now, using Ohm’s law V = IR, we have
𝐼=𝑉/𝑅 = 220 𝑉/31.25 Ω = 7.04 𝐴
Hence, the resistance of electric iron is 31.25 Ω and current through it is 7.04 A.
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Resistance of electric lamp, R1 = 100 Ω
Resistance of toaster, R2 = 50 Ω
Resistance of water filter, R3 = 500 Ω
Equivalent resistance Rp of the three appliances connected in parallel, is
1/100+1/50+1/500 =5+10+1/500 = 16/500 ⟹ 𝑅 = 500/16 =31.25 Ω
Resistance of electric iron = Equivalent resistance of the three appliances connected in parallel = 31.25 Ω
Applied voltage, V = 220 V
𝐼=𝑉/𝑅 = 220 𝑉/31.25 Ω = 7.04 𝐴
Given:
R1 = 100Ω, R2 = 50 Ω, R3 = 500 Ω all are connected in parallel
Potential difference = 220V
Thus, 1/R=1/100Ω+1/50Ω+1/500Ω=(5+10+1)/500Ω=16/500Ω
Therefore, R=500/16Ω=31.25Ω
Electric current (I) through the circuit = V/R
⇒I=220V/31.25Ω=7.04A
For electric iron
Since it takes as well current as three appliances, thus electric current through it = 7.04A
The electric current = 7.04 A and potential difference = 220 V
Thus, Resistance of electric iron = Total resistance of three appliances = 31.25 Ω
Thus, electric current through the electric iron = 7.04A
Resistance of electric iron = 31.25 Ω
When resistors are connected in parallel, the reciprocal of the equivalent resistance (Req) is equal to the sum of the reciprocals of the individual resistances. The formula for resistances in parallel is given by:
1/Req = 1/R₁ + 1/R₂ + 1/R₃ + . . .
For the given situation:
1/Req = 1/100 + 1/50 + 1/500
Now, let’s calculate Req:
1/Req = 1/100 + 2/100 + 1/500
1/Req = 3/100 + 1/500
1/Req = 15/500 + 1/500
1/Req = 16/500
Req = 500/16
Req = 31.25Ω
So, the equivalent resistance of the electric lamp, toaster, and water filter connected in parallel is 31.25Ω.
Now, if the electric iron is to draw the same current as all three appliances combined, we can use Ohm’s Law ( V= I ⋅R) to find the current (I) through the equivalent resistance:
I = V /Req
I = 220/31.25
I ≈7.04A
Therefore, the resistance of the electric iron connected to the same source is 31.25Ω, and the current through it is approximately 7.04A.