NCERT Solution for Class 10 Science Chapter 12
Electricity
NCERT Books for Session 2022-2023
CBSE Board and UP Board
Intext Questions
Page No-218
Questions No-3
An electric iron of resistance 20 ohm takes a current of 5 A. Calculate the heat developed in 30 s.
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According to Joule’s law of heating, the amount of heat produced is given by
H = VIt
Where,
V = IR = 5A × 20 Ω = 100 V
I = 5 A
and t = 30 seconds
So, 𝐻=100×5×30 𝐽
=15000 𝐽
=1.5×10⁴ 𝐽
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Here, R = 20 Ω, i = 5 A, t = 3s
Heat developed, H = I2 R t = 25 x 20 x 30 = 15,000 J = 1.5 x 104 J
Given, Electric current (I) = 5A, Resistance (R) = 20 Ω, Time (t) = 30s
Therefore, Heat produced (H) =?
Since, V=I×R=5A×20Ω=100V
We know that, Heat produced (H) = VIt
⇒H=100V×5A×30s=15000J=1.5×104J
The heat (Q) developed in an electric circuit can be calculated using the formula:
Q = I² . R . t
where:
» Q is the heat developed,
» I is the cerrent,
» R is the resistance, and
» t is the time.
Given:
I = 5A(current)
R = 20Ω (resistance)
t = 30s (time)
Substitude these values into the formula;
Q = (5A)² . (20 Ω) . (30s)
Q = 25 A² . 20Ω . 30s
Q = 25 . 20 . 30 J
Q = 15,000 J
Therefore, the heat developed in the electric iron in 30 seconds is 15,000 joules.