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$$
\begin{array}{l}{\text { For the circuit shown in Fig. E26.7 find the reading of the }} \\ {\text { idealized ammeter if the battery has an internal resistance of } 3.26 \Omega .}\end{array}
$$

$$
\frac{1}{R_{p}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}
$$

$$
=\frac{1}{45}+\frac{1}{15} \longrightarrow R_{p}=11.25 \Omega
$$

\(∴ Total \ Eq. R \Rightarrow R_{q}=18+11 \cdot 25+3 \cdot 26=32 \cdot 5 \Omega \)

$$
V=I R \rightarrow I=\frac{V}{R}
$$

$$
I=\frac{25}{32.5}=0.769 \mathrm{A}
$$

$$
\begin{array}{l}{\text { In Fig. } \mathrm{E} 26.11, R_{1}=} \\ {3.00 \Omega, \quad R_{2}=6.00 \Omega, \quad \text { and }} \\ {R_{3}=5.00 \Omega \text { . The battery has }} \\ {\text { negligible internal resistance. }} \\ {\text { The current } I_{2} \text { through } R_{2} \text { is }} \\ {4.00 \text { A. (a) What are the cur- }} \\ {\text { rents } I_{1} \text { and } I_{3} ? \text { (b) What is the }} \\ {\text { emf of the battery? }}\end{array}
$$

(a) $$
V_{2} = I_{2} R_{2} \quad = 4 * 6=24 \mathrm{v}
$$

$$
V_{1}=V_{2}=24 V
$$

$$
\rightarrow I_{1}=\frac{V_{1}}{R_{1}}=\frac{24}{3}=8 A
$$

$$
I_{3}=I_{1}+I_{2}=4+8=12 A
$$

(b) $$
\varepsilon=V_{1}+V_{3}
$$

$$
V_{3}=I_{3}* R_{3}
$$$$
=12 * 5=60 \mathrm{V}
$$

$$
=24+60=84 \mathrm{v}
$$

$$
\begin{array}{l}{\text { Compute the equivalent resistance }} \\ {\text { of the network in Fig. } \mathrm{E} 26.13, \text { and find the }} \\ {\text { current in each resistor. The battery has }} \\ {\text { negligible internal resistance. }}\end{array}
$$

$$
\frac{1}{R_{12}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}=
$$$$
\frac{1}{3}+\frac{1}{6}
$$

$$
R_{12}=\frac{R_{1} R_{2}}{R_{1}+R_{2}}=
$$$$
\frac{3 * 6}{9}=2 \Omega
$$

$$
\frac{1}{R_{34}}=\frac{1}{R_{3}}+\frac{1}{R_{4}}=
$$$$
\frac{1}{12}+\frac{1}{4}
$$

$$
R_{34}=\frac{R_{3} R_{4}}{R_{3}+R_{4}}=
$$$$
\frac{12 * 4}{16}=3 \Omega
$$

$$
R e q=2+3
$$

$$
R e q=R_{12}+R_{34}
$$

$$
=5 \Omega
$$

$$
I=\frac{V}{R}=\frac{60}{5}=129
$$

$$
V_{12}=I R_{12} =(12)(2)=24 \mathrm{V}
$$

$$
V_{34}=I R_{34}=(12)(3)=36 \mathrm{V}
$$

$$
I_{1}=\frac{V_{12}}{R_{1}} = \quad \frac{24}{3}=8 A
$$

$$
I_{2}=\frac{V_{12}}{R_{2}}=\frac{24}{6}=4 A
$$

$$
I_{3}=\frac{V_{34}}{R_{3}}=\frac{36}{12}=3 A
$$

$$
I_{4}=\frac{V_{34}}{R_{4}}=\frac{36}{4}=9 A
$$

 

 

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