The current in Fig. 26-1 is 0.125 A in the direction shown....

The current in Fig. $26-1$ is $0.125$ A in the direction shown. For each of the following pairs of points, what is their potential difference, and which point is at the higher potential? (a) $A$, $B$; $($ b) $B, C ;(c) C$, $D ;(d) D, E ;(e) C, E ;(f) E, C .$ Recall the following facts: (1) The current is the same $(0.125 \mathrm{~A})$ at all points in this circuit because the charge has no other place to flow. (2) Current always flows from high to low potential through a resistor. (3) The positive terminal of a pure emf (the long side of its symbol) is always the high-potential terminal. Mark the long sides of the batteries with plus signs ( $+$ ) and the short sides with minus signs ( $-$ ). Current streams out of the positive terminal of the 12-V battery and, in this case, flows clockwise around the circuit because the 12-V battery dominates over the 9.0-V battery. For each resistor place a + on the side where current enters and a where it leaves. When current passes through a resistor from + to - it experiences what is called a "voltage drop." Taking potential drops as negative: (a) $V_{A B}=-I R=-(0.125$ A) $(10.0 \Omega)=-1.25 \mathrm{~V} ; A$ is higher. (b) $V_{B C}=-\varepsilon=-9.00 \mathrm{~V} ; B$ is higher. (c) $V_{C D}=-(0.125 \mathrm{~A})(5.00 \Omega)-(0.125 \mathrm{~A})(6.00 \Omega)=-1.38 \mathrm{~V} ; C$ is higher. (d) $V_{D E}=+\varepsilon=+12.0 \mathrm{~V} ; E$ is higher. (e) $V_{C E}=-(0.125 \mathrm{~A})(5.00 \Omega)-(0.125 \mathrm{~A})(6.00 \Omega)+12.0 \mathrm{~V}=$ $$ \begin{array}{l} +10.6 \mathrm{~V} ; E \text { is higher. }\\ \text { (f) } V_{E C}=-(0.125 \mathrm{~A})(3.00 \Omega)-(0.125 \mathrm{~A})(10.0 \Omega)-9.00 \mathrm{~V}=\\ -10.6 \mathrm{~V} \text { ; } E \text { is higher. }\\ \text { Notice that the answers to }(e) \text { and }(f) \text { agree with each other. } \end{array} $$

The current in Fig. $26-1$ is $0.125$ A in the direction shown. For each of the following pairs of points, what is their potential difference, and which point is at the higher potential? (a) $A$, $B$; $($ b) $B, C ;(c) C$, $D ;(d) D, E ;(e) C, E ;(f) E, C .$
Recall the following facts: (1) The current is the same $(0.125 \mathrm{~A})$ at all points in this circuit because the charge has no other place to flow. (2) Current always flows from high to low potential through a resistor. (3) The positive terminal of a pure emf (the long side of its symbol) is always the high-potential terminal. Mark the long sides of the batteries with plus signs ( $+$ ) and the short sides with minus signs ( $-$ ). Current streams out of the positive terminal of the 12-V battery and, in this case, flows clockwise around the circuit because the 12-V battery dominates over the 9.0-V battery. For each resistor place a + on the side where current enters and a where it leaves. When current passes through a resistor from + to
- it experiences what is called a "voltage drop." Taking potential drops as negative:
(a) $V_{A B}=-I R=-(0.125$ A) $(10.0 \Omega)=-1.25 \mathrm{~V} ; A$ is higher.
(b) $V_{B C}=-\varepsilon=-9.00 \mathrm{~V} ; B$ is higher.
(c) $V_{C D}=-(0.125 \mathrm{~A})(5.00 \Omega)-(0.125 \mathrm{~A})(6.00 \Omega)=-1.38 \mathrm{~V} ; C$ is
higher.
(d) $V_{D E}=+\varepsilon=+12.0 \mathrm{~V} ; E$ is higher.
(e) $V_{C E}=-(0.125 \mathrm{~A})(5.00 \Omega)-(0.125 \mathrm{~A})(6.00 \Omega)+12.0 \mathrm{~V}=$ $$
\begin{array}{l}
+10.6 \mathrm{~V} ; E \text { is higher. }\\
\text { (f) } V_{E C}=-(0.125 \mathrm{~A})(3.00 \Omega)-(0.125 \mathrm{~A})(10.0 \Omega)-9.00 \mathrm{~V}=\\
-10.6 \mathrm{~V} \text { ; } E \text { is higher. }\\
\text { Notice that the answers to }(e) \text { and }(f) \text { agree with each other. }
\end{array}
$$

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