Question

Consider the circuit of Fig. E25.30. (a)What is the total rate at which electrical energy is dissipated in the 5.0-Ω and 9.0-Ω resistors? (b) What is the power output of the 16.0-V battery? (c) At what rate is electrical energy being converted to other forms in the 8.0-V battery? (d) Show that the power output of the 16.0-V battery equals the overall rate of consumption of electrical energy in the rest of the circuit.

Fig. E25.30.

Short answer

1. The total rate at which electrical energy is dissipated in 5.0 Ω and 9.0 Ω is.
2. The power output of the 16.0 V battery is 7.2 W
3. The rate at which electrical energy is converted to other forms in the 8.0 V battery is 4.1 W
4. It is shown that the power output of the 16.0-V battery equals the overall rate of consumption of electrical energy in the rest of the circuit.

Step-by-step solution

(a) Determination of the total rate at which electrical energy is dissipated in 5.0 Ω and 9.0 Ω.

Expression of power dissipation in a resistor of resistance R is,

$P=l^{2}R$

Solve for current using Ohm’s Law,

$$\begin{gathered} l=\frac{V}{R} \\ =\frac{8.0V}{17\Omega } \\ =0.47A \end{gathered}$$

Use the value of I in the power expression,

$$\begin{gathered} {\mathrm{P}}_{5\mathrm{\Omega }}={\mathrm{l}}^2\mathrm{R} \\ ={(0.47\mathrm{A})}^2\left(5.0\mathrm{\Omega }\right) \\ =1.1\mathrm{W} \\ \mathrm{Similarly}, \\ {\mathrm{P}}_{9\mathrm{\Omega }}={\mathrm{l}}^2\mathrm{R} \\ ={(0.47\mathrm{A})}^2\left(9.0\mathrm{\Omega }\right) \\ =2.0\mathrm{W} \\ \mathrm{Thus},\mathrm{the}\mathrm{total}\mathrm{power}\mathrm{is}, \\ \mathrm{P}={\mathrm{P}}_{5\mathrm{\Omega }}+{\mathrm{P}}_{9\mathrm{\Omega }} \\ =1.1\mathrm{W}+2.0\mathrm{W} \\ =3.1\mathrm{W} \end{gathered}$$

(b) Determination of the power output of the 16.0 V batteries.

The power output is given by,

$$\begin{gathered} {\mathrm{P}}_{16\mathrm{V}}=\mathrm{\epsilon l}-{\mathrm{l}}^2\mathrm{r} \\ \mathrm{Substitute}\mathrm{\epsilon }=16\mathrm{V},\mathrm{l}=0.47\mathrm{A}\mathrm{and}\mathrm{r}=1.6\mathrm{\Omega } \\ {\mathrm{P}}_{16\mathrm{V}}=(16\mathrm{V})(0.47\mathrm{A})-{(0.47\mathrm{A})}^2(1.6\mathrm{\Omega }) \\ =7.2\mathrm{W} \end{gathered}$$

Thus, the power output of the 16.0 V battery is 7.2 W

(c) Determination of the rate at which electrical energy is converted to other forms in the 8.0 V batteries.

Rate of conversion of electrical energy is the power output and the expression for the 8.0 V battery is given as,

$$\begin{gathered} {\mathrm{P}}_{8\mathrm{V}}=\mathrm{\epsilon l}-{\mathrm{l}}^2\mathrm{r} \\ \mathrm{Substitute}\mathrm{\epsilon }=8.0\mathrm{V},\mathrm{l}=0.47\mathrm{A}\mathrm{and}\mathrm{r}=1.4\mathrm{\Omega } \\ {\mathrm{P}}_{8\mathrm{V}}=(8\mathrm{V})(0.47\mathrm{A})-{(0.47\mathrm{A})}^2(1.4\mathrm{\Omega }) \\ =4.1\mathrm{W} \end{gathered}$$

Thus, the power output of the 8.0 V battery is 4.1 W.

(d) Proof that the power output of the 16.0-V battery equals the overall rate of consumption of electrical energy in the rest of the circuit.    

Infer from part (a), (b) and (c)

It is shown that the power output for 16.0 V in part (b) is equal to the power dissipated for 8.0 V battery in part (c) and the combined power loss across 5.0Ω and 9.0 Ω resistor in part (a).