The battery in Fig. E26.28 is removed from the circuit and replaced by a battery, with its negative terminalnext to point b. The rest of the circuit is as shown in the figure. Find (a) the current in each branch and (b) the potential differenceof point a relative to point b.

Potential difference between any two points is defined as the amount of work done in moving a unit charge from one point to another

Solutlon
(a) When 5 V battery is replaced by 15 V, the current source will be 15 V battery Which supplies 10 V and 10 ohms and want to find the current in each branch- The current directions will change; the current in the top branch is I1 (left to right), in the
middle branch is I2 (right to left) and in the bottom branch is I3 (left to right )- apply the loop rule to get the variables Where the loop rule is a statement that the electrostatic force is conservative
Suppose go around a loop, measuring potential differences across circuit elements as We go and the algebraic sum ofthese differences is zero when We return to the starting point The figure below shows the loop directions and the paths thatl take to get the target variables-
use loop 1 (Closed blue path) and apply equation 26.6 as shOWn in the figure below where the direction of our travel iscounterclockwise
$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ 2{\mathrm{l}}_1+10\mathrm{V}+3{\mathrm{l}}_1+1{\mathrm{l}}_2+4{\mathrm{l}}_2-15\mathrm{V}=0 \end{gathered}$$
V is negative because the direction of traveling is from positive to negative terminal in the battery (See ?gure 26.8a )-
The terms are positive because the traveling direction is the same direction of the
Now
Ne can solve the summation to get the next equation

${\mathrm{l}}_1+{\mathrm{l}}_2=1\mathrm{A}$

With the same steps let us use loop 2 (Closed black path) and apply equation 26.6 as shown in the figure below where the direction of our travel is counterclockwise

$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ 4{\mathrm{l}}_2+15\mathrm{V}-1{\mathrm{l}}_2-10{\mathrm{l}}_3=0 \end{gathered}$$

,V is positive because the direction of traveling is from negative to positive terminal in the battery (See ?gure 26.8a )-
The terms (4 ohms)I2, (1 ohms)I2 and (10 ohms)I3 are negative because the traveling direction is the same direction of the current
(See ?gure 26.8b ). Now We can solve the summation to get the next equation
${\mathrm{l}}_2+2{\mathrm{l}}_3=3\mathrm{A}$
The 15 V battery supplies the 10 V battery and resistance 10 9, so the current (I2) shows from 15 V battery splits at point (1 totwo current I1 and I3 apply the junction rule in this case. Where the junction rule is based on conservation of electric charge and the
current enters ajunction point is equal to the current ?OWS out from this point, so in our circuit, we could get the next as I2
$$\begin{gathered} {\mathrm{l}}_2={\mathrm{l}}_1+{\mathrm{l}}_3 \\ {\mathrm{l}}_1={\mathrm{l}}_2-{\mathrm{l}}_3 \\ {\mathrm{l}}_1+{\mathrm{l}}_2=1\mathrm{A} \end{gathered}$$
Now We have three equations with three variables- Let us plug the expression of I1 from equation (3) into equation (1), henc
get a new form of equation (1) as next

$$\begin{gathered} {\mathrm{l}}_2+{\mathrm{l}}_1+{\mathrm{l}}_3 \\ {\mathrm{l}}_1={\mathrm{l}}_2-{\mathrm{l}}_3 \\ {\mathrm{l}}_1+{\mathrm{l}}_2+1\mathrm{A} \\ 2{\mathrm{l}}_2-{\mathrm{l}}_3=1\mathrm{A} \\ + \\ -1{\mathrm{l}}_2+2{\mathrm{l}}_3=3\mathrm{A} \\ 5/2{\mathrm{l}}_3=5/2\mathrm{A} \\ {\mathrm{l}}_3=1\mathrm{A} \end{gathered}$$

Thisisthecurrentflowsinthebottombranch1A

Now We can plug the value for I3 into equation (2) to get I2 as next

$$\begin{gathered} {\mathrm{l}}_2=3\mathrm{A}-2{\mathrm{l}}_3 \\ =1\mathrm{A} \end{gathered}$$

This is the current flOWS in the middle branch1A

Again, let us plug the values for 12 and I3 into equation (3) to get I1

$$\begin{gathered} {\mathrm{l}}_1={\mathrm{l}}_2-{\mathrm{l}}_3=1\mathrm{A}-1\mathrm{A} \\ =0 \end{gathered}$$

No current in the top branch-0A

(b)tofindthepotentialdifferencebetweenthepointsa.andb.Thepotentialbetweena,andbisthesumofthepotentialdropinthepathab.50letusstartfroma,tobandtakethepathwiththeredcolorasshoWninthe?gurebelow
where, in this path, We have No resistors 3.00 D and 4.00 9 so Val, will be
$$\begin{gathered} {\mathrm{V}}_{\mathrm{ab}}={\mathrm{V}}_{\mathrm{b}}-{\mathrm{V}}_{\mathrm{a}} \\ =-3{\mathrm{l}}_1-4{\mathrm{l}}_2 \\ =-4\mathrm{V} \end{gathered}$$
Don't forget that the terms (3.00 ohms) and (4.00 ohms) are negative because the traveling path that we take, is in the sarr
direction of the current flows in both resistors as shOWn in the ?gure below. As the potential is negative, therefore, at poin‘
the potential is higher than at point b.

Therefore the potential differnce is = 4.0 V and the potential at a is higher than b-