91Ó°ÊÓ

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 $\left(\mathrm{Î\copyright }\right)$and $V$is the potential difference volt $\left(V\right)$.

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

So, formula for the terminal potential difference is:

$V_{ab}=\mathrm{Î\mu }-Ir$

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

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

Where, $\mathrm{ÒÏ}$is resistivity $\mathrm{Î\copyright }·\mathrm{m},L$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,

$$\begin{gathered} R=5\mathrm{Î\copyright } \\ r=1\mathrm{Î\copyright } \\ \mathrm{Î\mu }=12\mathrm{V} \end{gathered}$$

The rate of energy conversion:

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

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 .