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$$
\begin{array}{l}{\text { Consider the circuit shown in Fig. E25.28. The termi- }} \\ {\text { nal voltage of the } 24.0-\mathrm{V} \text { battery is } 21.2 \mathrm{V} . \text { What are (a) the inter- }} \\ {\text { nal resistance } r \text { of the battery and }(\mathrm{b}) \text { the resistance } R \text { of the circuit }} \\ {\text { resistor? }}\end{array}
$$

$$
\varepsilon=24 v \quad,\quad I=4 A \quad, \quad V_{a b}=21 \cdot 2 v
$$

(?) $$
R??
$$

(a) $$
V_{a b}=\varepsilon-I r \Rightarrow r=
$$$$
\frac{\varepsilon-V_{a b}}{I}=\frac{24-21 \cdot 2}{4}
$$

\(∴ r=0.7 \Omega \)

(b) $$
V_{a b}-I R=0
$$$$
\longrightarrow R=\frac{V_{a b}}{I}=\frac{21.2}{4}=5.3 \Omega
$$

$$
R+r=6 \Omega
$$

$$
I=\frac{V}{R} \longrightarrow \frac{24}{6}=4 A
$$

$$
\begin{array}{l}{\text { A copper transmission cable } 100 \mathrm{km} \text { long and } 10.0 \mathrm{cm} \text { in diameter carries a current of } 125 \mathrm{A} \text { . }} \\ {\text { (a) What is the potential drop across the cable? (b) How much electrical energy is dissipated as }} \\ {\text { thermal energy every hour? }}\end{array}
$$

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

$$
P=V I
$$                $$
\text { energy } = P.t
$$

(a) $$
\rho=1.72 * 10^{-8} \Omega \cdot m
$$        $$
A=\pi r^{2} \quad r=0.05 m
$$

$$
R=\frac{\rho L}{A}=\frac{1.72 * 10^{-8} * 100 * 10^{3}}{\pi(0.05)^{2}}=0.219 \Omega
$$

$$
V=I R=125 * 0.219=27.4 \mathrm{V}
$$

(b) $$
p=V I=27.4 * 125=3422 w=3422  \ \mathrm{J/s}
$$

\(∴ \text { energy } =P t \quad=3422 * 3600=1.23*10^7 J \)

$$
\begin{array}{l}{\text { When switch } \mathrm{S} \text { in Fig. E25.33 }} \\ {\text { is open, the voltmeter } \mathrm{V} \text { of the battery }} \\ {\text { reads } 3.08 \mathrm{V} \text { . When the switch is closed, }} \\ {\text { the voltmeter reading drops to } 2.977 \mathrm{V} \text { , }} \\ {\text { and the ammeter A reads } 1.65 \mathrm{A} \text { . Find }} \\ {\text { the emf, the internal resistance of the }} \\ {\text { battery, and the circuit resistance of the }} \\ {\text { Assume that the two meters are ideal, }} \\ {\text { so they don't affect the circuit. }}\end{array}
$$

Open

$$
I=0
$$

$$
V_{a b}=\varepsilon=3.08 V
$$

Closed

$$
V_{a b}=\varepsilon- I{r}=2.97 V
$$

$$
r=\frac{\varepsilon-2.97}{I}
$$

$$
=\frac{3 \cdot 08-2.97}{1.65}
$$$$
=0.067 \Omega
$$

$$
V_{ a b}=I R
$$$$
\rightarrow {R}=
$$$$
\frac{V_{a b}}{I}
$$

$$
=\frac{2.97}{1.65}=1.8 \Omega
$$

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