Question

find the heat energy liberated when 20 A electric current passing through a resistor of 10 ohm for 10 minutes. what will be the change in heat if the electric current is doubled?
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Answer 1

Solution :-

As per the given data ,

• Current ( I) = 20 A
• Resistor (R) = 10 Ω
• Time taken (t) = 10 min = 10 x 60 = 600 s

Heat energy liberated is given by the formula ,

➜ H = I ² Rt

Now let's substitute the given values in the above equation ,

➜ H = 20 x 20 x 10 x 600

➜ H = 24,00,000 J

Now the current is doubled ( I = 40 A )

➜ H' = I '² Rt

Now let's substitute the given values in the above equation ,

➜ H '= 40 x 40 x 10 x 600

➜ H '= 96,00,000 J

Change in heat (ΔH)

➜ ΔH = H' - H

➜ ΔH = 96,00,000 - 24,00,000

➜ ΔH = 74,00,000J

The change in heat energy is 74,00,000 J

Answer 2

Given :

Current, I = 20 A

Resistance, R = 10 Ω

Time taken, t = 10min = 600sec

To find:

the change in heat if the electric current is doubled

Solution :

here, we can use Joules law of heating,

According to joules law,

The heat produced in a resistor is directly proportional to

• the square of current for a given resistance ( H ∝ I²)
• the resistance for a given current (H ∝ R)
• the time for which the current flows through the resistor (H ∝ t).

.i.e.,

H = I²Rt

here, H denotes heat, I denotes Current, R denotes resistance and t denotes time

So, using joules law of heating,

➝ H₁ = I ² Rt

now, by substituting all the given values in the formula

➝ H₁ = (30)²(10)(600)

➝ H₁ = 24,00,000 J

now, If the current in doubled then I₁ = 40A.

➝ H₂ = I₁² Rt

by substituting all the value in the formula,

➝ H₂ = (40)²(10)(600)

➝ H₂ = 96,00,000 J

theChange in heat

➠ ΔH = H₂ - H₁

➠ ΔH = 96,00,000 - 24,00,000

➠ ΔH = 74,00,000J = 7.4 × 10⁶J

thus, the change in heat energy is 7.4 × 10⁶J$.$

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