In the circuit in Fig. E25.47, find (a) the rate of conversion of internal (chemical) energy to electrical energy within the battery; (b) the rate of dissipation of electrical energy in the battery; (c) the rate of dissipation of electrical energy in the external resistor.

According to Ohm’s law, the current flowing through the conductor is directly proportional to the voltage across the two points.

$V=IR$

Where, $I$is current in ampere $\left(A\right)$, $R$is resistance in ohms $\left(\Omega \right)$and $V$is the potential difference volt $\left(V\right)$.

If the real source of emf $\epsilon$has internal energy$r$, then its terminal potential difference$V_{ab}$depends upon the current.

So, formula for the terminal potential difference is:

$V_{ab}=\epsilon -Ir$

The ratio of V toI for a particular conductor is called its resistance R

$R=\frac{V}{I}or\frac{\rho L}{A}$

Where, $\rho$is resistivity $\mathrm{\Omega }·\mathrm{m},L$is length in m and $A$is area in ${\mathrm{m}}^2$.

The power role="math" localid="1655721678088" $\mathit{\left(}\mathit{P}\mathit{\right)}$is the product of potential difference $\mathit{\left(}\mathit{V}\mathit{\right)}$and the current $\mathit{\left(}\mathit{I}\mathit{\right)}$.

$P=VIor{I}^{2}Ror\frac{V^2}{R}$

Given that,

$$\begin{gathered} R=5\mathrm{\Omega } \\ r=1\mathrm{\Omega } \\ \epsilon =12\mathrm{V} \end{gathered}$$

The rate of energy conversion:

$$\begin{gathered} P=\epsilon I \\ =\epsilon \left(\frac{\epsilon }{R+r}\right)\left[\because I=\frac{\epsilon }{R+r}\right] \\ =\left(12\right)\left(\frac{12}{5+1}\right) \\ =24\mathrm{W} \end{gathered}$$

Hence, the rate of conversion of internal (chemical) energy to electrical energy within the battery is 24 W .

The rate of dissipation of electrical energy:

$$\begin{gathered} P=I^{2}r \\ ={\left(2\right)}^2\left(1\right) \\ =4\mathrm{W} \end{gathered}$$

Hence, the rate of dissipation of electrical energy in the battery is 4 W .

The rate of dissipation of electrical energy:

$$\begin{gathered} P=I^{2}R \\ ={\left(2\right)}^2\left(5\right) \\ =20\mathrm{W} \end{gathered}$$

Hence, rate of dissipation of electrical energy in the external resistor is 20 W .