 # Explain the heating effect of electric current

HEATING EFFECT OF ELECTRIC CURRENT: The phenomenon when a steady current flows through conductor the energy liberated during the current flow exhibited as heat

Consider a conductor of AB having potential difference V, potential at A is Va and potential at B is VB

If VA > VB

The potential energy of an electron at A is

UA = – q VA

Similarly

Potential energy of an electron at B is

UB = – q VB

If current flow through the conductor is (i) then amount of charge flows through the conductor in dt time is

dq = i dt

Number of electron flowing from end B to A in dt time is

$$\begin{array}{l}\frac{q}{e}=\frac{Idt}{e}\end{array}$$

When electron moves from lower potential to higher potential there will be energy loss of electron (i.e., loss in potential energy)

$$\begin{array}{l}dw=\left( \frac{Idt}{e} \right)\left[ \left( -e{{V}_{B}} \right)-\left( -e{{V}_{A}} \right) \right]\end{array}$$

$$\begin{array}{l}=\frac{Idt\,\,e}{e}\left( {{V}_{A}}-{{V}_{B}} \right)\end{array}$$

$$\begin{array}{l}dw=I\,dt\,v\end{array}$$

$$\begin{array}{l}dw=V\,I\,dt\end{array}$$
….. (1)

Electrical power (p) is defined as

$$\begin{array}{l}P=\frac{dW}{dt}=\frac{V\,I\,dt}{dt}\end{array}$$

P = V I ….. (2)

V = I R

P = V I = I2R

In general if electric current flows for a time t, the energy liberated will be

$$\begin{array}{l}H=\frac{w}{J}\end{array}$$

$$\begin{array}{l}J=4.2\,\,J/cal\end{array}$$

$$\begin{array}{l}H=\frac{w}{4.2}=\frac{VIt\,}{4.2}\end{array}$$

$$\begin{array}{l}H=\frac{{{I}^{2}}Rt}{4.2}\end{array}$$

Alternatively

When a steady current (i) flows for time (t) through a conductor by a source of emf E. then amount of charge that flow through conductor in time (t) is

Q = I t

Electrical energy delivered to charge flow through the conductor is

W = Q V = V I t

Thus, electric power given to the circuit is

$$\begin{array}{l}P=\frac{W}{t}=\frac{VIt}{t}=VI\end{array}$$

$$\begin{array}{l}P=VI={{I}^{2}}R\end{array}$$

Let as consider a simple circuit:

In the given circuit.

The rate at which chemical energy is converted into electrical energy is EI and power delivered across the external resistor R is I2R and power dissipated across the internal resistance of the cell is i2r, it can written as

$$\begin{array}{l}EI={{I}^{2}}R+{{I}^{2}}r\end{array}$$

Since,

When a steady current flows through conductor, heat is produced in it. This is known as Joule’s Heating effect the heat developed across the external load resistance is given by

H = I2Rt (2) (1)