The current in Fig. $26-1$ is $0.125 \mathrm{~A}$ 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 $12-\mathrm{V}$ battery and, in this case, flows clockwise around the circuit because the $12-\mathrm{V}$ battery dominates over the $9.0-\mathrm{V}$ battery. For each resistor place $\mathrm{a}+$ 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) $V_{A B}=-I R=-(0.125 \mathrm{~A})(10.0 \Omega)=-1.25 \mathrm{~V} ; A$ is higher.
(b) $V_{B C}=-\ell=-9.00 \mathrm{~V} ; B$ 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) $V_{D E}=+\mathscr{C}=+12.0 \mathrm{~V} ; E$ is higher.
(e) $V_{C E}=-(0.125 \mathrm{~A})(5.00 \Omega)-(0.125 \mathrm{~A})(6.00 \Omega)+12.0 \mathrm{~V}=+10.6 \mathrm{~V} ; E$ is higher.
(f) $V_{E C}=-(0.125 \mathrm{~A})(3.00 \Omega)-(0.125 \mathrm{~A})(10.0 \Omega)-9.00 \mathrm{~V}=-10.6 \mathrm{~V} ; E$ is higher.
Notice that the answers to $(e)$ and $(f)$ agree with each other.