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**Derive the expression** for the heat produced due to a current â€˜Iâ€™ flowing for a time interval â€˜tâ€™ through a resistor â€˜Râ€™ having a potential difference â€˜Vâ€™ across its ends. With which name is the relation known? How much heat will an instrument of 12 W produce in one minute if it is connected to a battery of 12 V?

Since a conductor provides resistance to the flow of current, some work must be done by the current continuously to keep itself flowing.

When an electric charge $Q$ moves against a potential difference $V$, the amount of work done is given as-

$W=Q\times V$ ------------(1)

We know, the electric current, $I$ is given as-

$I=\frac {Q}{t}$

Then, in terms of $Q$, it is given as-

$Q=I\times t$ ------------(2)

By ohm’s law, we know that-

$R=\frac {V}{I}$

Then, in terms of $V$, it is given as-

$V=I\times R$ ------------(3)

Now, putting equation (2) and (3) in eq, (1) we get-

$W=I\times t\times I\times R$

$W=I^{2}Rt$

Assuming that all the electrical work done is converted into heat energy, we get-

Heat produced, $H$ = Work done in the above equation

Thus, $H=I^{2}Rt\ Joules$

This relation is known as **Joule’s law of heating.**

**Given: **

Power of instrument, $P=12W$

Voltage, $V=12V$

Time, $t=1minute=60sec$

**To find: **Heat produced by the instrument, $H$.

**Solution:**

We know that the formula of electric power is given as-

$P=V\times I$

Substituting the given values we get-

$12=12\times I$

$I=\frac {12}{12}$

**
**

**
**

Thus, the current flowing in the instrument is **1 Ampere.**

From Ohm's law we know that-

$V=I\times R$

Substituting the value of $I$ and $R$ we get-

$12=1\times R$

$R=\frac {12}{1}$

$R=12\Omega$

Thus, the resistance of the instrument is **12 Ohm.**

From the formula of heating we know that-

$H=I^{2}Rt$

Substituting the required values we get-

$H=1^{2}\times 12\times 60$

$H=720J$

Thus, the heat produced by the instrument is **720 Joules.**