Question

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.

Short answer

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

(b) The rate of dissipation of electrical energy in the battery is 4 W .

(c) The rate of dissipation of electrical energy in the external resistor is 20 W .

Step-by-step solution

Define the ohm’s law, resistance (R) and power (P) .

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 $\mathit{\left(}\mathit{I}\mathit{\right)}$.

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

Determine the rate of conversion of internal energy to electrical energy.

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 .

Determine the rate of dissipation of electrical energy in the battery.

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 .

Determine the rate of dissipation of electrical energy in the external resistor.

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 .