Energy, Power And Heating Effect

When a current I flows for time t from a source of emf E, then the amount of charge that flows in time t is Q = It.

Electrical energy delivered W = Q.V = VIt

Thus, Power given to the circuit, = W/t = VI or V²/R or I²R

In the circuit

E. I = I²R + I² r, where

EI is the rate at which chemical energy is converted to electrical energy, I²R is power supplied to the external resistance R and I²r is the power dissipated in the internal resistance of the battery.

An electrical current flowing through conductor produces heat in it. This is known as Joule's effect. The heat developed is given by H = I².R.t joule,

where I = current in ampere , R = resistance in $\Omega$, t = time in second.

Maximum Power Theorem

In a circuit, for what value of the external resistance the maximum power be drawn from a battery? For the shown network power developed in resistance R equals

$P=\frac{E^2.R}{{(R+r)}^2}$ ($\because$ I =$\frac{E}{R+r}$ and P = I²R )

Now, for dP/dR = 0 (for P to be maximum $\frac{dP}{dR}=0$)

$\Rightarrow E^2.\frac{{(R+r)}^2-2(R)(R+r)}{{(R+r)}^4}=0$

$\Rightarrow$ R + r = 2R $\Rightarrow$ R = r

$\Rightarrow$ The power output is maximum, when the external resistance equals the internal resistance.

R = r

Example : A copper wire having a cross-sectional area of 0.5 mm² and a length of 0.1 m is initially at 25^oC and is thermally insulated from the surroundings. If a current of 10 A is set up in this wire

(a) Find the time in which the wire starts melting. The change of resistance of the wire with temperature may be neglected.

(b) What will this time be, if the length of the wire is doubled?

Density of Cu = 9 ´ 10³ Kg m^-3 specific heat of Cu = 9 ´ 10^-2 Cal Kg^{-1 o}C^-1, M.P. (Cu) 1075 ^oC and specific resistance = 1.6 ´ 10^-8$\Omega$m.

Solution : (a) Mass of Cu = Volume x density

= 0.5 x 10^-6 x 0.1 x 9 x 10³ = 45 x 10^-5 Kg.

Rise in temperature = $\theta$ = 1075-25 = 1050 ^oC.

Specific heat = 9 x 10^-2 Kg^{-1 o}C x 4.2 J

$\Rightarrow$ I²Rt = mSq $\Rightarrow t=\frac{mS\theta }{I^2.R}$

\begin{align}  butR=\dfrac{\rho L}{A}=\dfrac{1.6\,\times \,{{10}^{-8}}\,\times \,\,0.1}{0.5\,x\,{{10}^{-6}}}=3.2\times \,{{10}^{-3}}\,\Omega \\ \Rightarrow \,\,t=\dfrac{45\,\times \,4.2\,\times \,{{10}^{-5}}\,\times \,1050\,\times \,0.09}{10\,\times \,10\,\times \,3.2\,\times \,{{10}^{-3}}}=558\,s \\\end{align}

(b) When the length of wire is doubled, R is doubled, but correspondingly mass is also doubled. Therefore, wire will start melting in the same time.