Question

The heat produced by electric current is given by the relationship, H = l2Rt. How does H vary
when (a) current (I) is doubled? (b) resistance (R) is halved? (c) time (t) is doubled?

Answer 1

Explanation:

The heating effect of an electric current depends on three factors:

The resistance, R of the conductor. A higher resistance produces more heat.

The time, t for which current flows. The longer the time the larger the amount of heat produced.

The amount of current,

Answer 2

Given:

Heat energy produced in the conductor $=I^{2} RT$

To find:

Variation of heat.

Solution:

• The Joule's law of heating says that the heat energy produced in the conductor is directly proportional to the square of electric current, resistance and the time till which the current is supplied.

                             $H=I^{2} RT$

Now, according to the question,

• When current is doubled.    

The new current becomes  $I'=2I$

$H'=I'^{2} RT$

$H'=(2I)^{2} RT$

$H'=4I^{2} RT$

$H'=4H$

Hence, on doubling the current, the heat produced becomes 4 times the initial value.

• When resistance (R) is halved.

The new resistance becomes $R'=\frac{R}{2}$

$H'=I^{2} R'T$

$H'=I^{2} (\frac{R}{2} )T$

$H'=\frac{I^{2} RT}{2}$

$H'=\frac{H}{2}$

Hence, if we reduce the resistance by halve, the new resistance becomes halved.

• When time (T) is halved.

The new time becomes $T'=\frac{T}{2}$, and the energy becomes,

$H'=I^{2} RT'$

$H'=I^{2} R(\frac{T}{2} )$

$H'=\frac{I^{2} RT}{2}$

$H'=\frac{H}{2}$

Hence, reducing the time by 2, the new heat energy becomes half of the original value.

Final answer:

Hence,

1. Increasing the current by 2 times, the heat energy becomes 4 times.
2. Reducing the resistance by half, the heat energy becomes half of the original value.
3. Reducing the time by half, the heat energy becomes half of the original value.