Question

Thebattery in Fig. E26.28 is removed from the circuit and reinserted with the opposite polarity, so that its positive terminal is now next to point a. The rest of the circuit is as shown in the figure. Find (a) the current in each branch

and (b) the potential difference Vabof point arelative to point b.

Short answer

thecurrentflOWsinthebottombranch-0.2A

the cureent in middle branch is 1.4A

the current in the upper branch is 1.6A

the potential diffeernce at a is higher than b and equal to 10.4V

Step-by-step solution

Step 1:About polarity

Polarity is defined as the property in a molecule, or compound through which they are either attracted or repelled by an electric charge because of an asymmetrical arrangement of electropositive or electronegative atoms around the center of the species

Determine the the current in each branch

SOlUtlon
(a)solvefor the current in each branch when the 10 V change its polarity.

The current directions will change

; Thecurrent in the top branch is (left to right), in the middle branch is (right to left) and in the bottom branch is (right toleft)- To find$1,{\mathrm{l}}_2,{\mathrm{l}}_3$

Apply the loop rule to get the variables where the loop rule is a statement that the electrostatic force is conservative-
Suppose 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 loop directions and the paths that
We will take to get the target variables use loop 1 (Closed blue path) and apply equation where the direction of our travel iscounterclockwise
$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ 2{\mathrm{l}}_1-10\mathrm{V}+3{\mathrm{l}}_1+1{\mathrm{l}}_{2}4{\mathrm{l}}_2-5\mathrm{V}=0 \end{gathered}$$
10 ,V and 5 ,V are negative because the direction of traveling is from positive to negative terminal in the battery

The terms (3 ohms)I1, (2 ohms)I1, (1 ohms)I2 and (4 ohms)I2 are positive because the traveling direction is the same direction of the
current in each resistance

Now we can solve the summation to get the next equation
${\mathrm{l}}_1+{\mathrm{l}}_2=3\mathrm{A}$

:Determine the current in upper ,middle and lower branch

Withthesamestepsuseloop2(Closedblackpath)andapplyequation26-6asshowninthefigurebelowwherethe directionofourtraveliscounterclockwise
$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ -4{\mathrm{l}}_2-5\mathrm{V}+1{\mathrm{l}}_2+10{\mathrm{l}}_2+10{\mathrm{l}}_3=0 \end{gathered}$$
5 ,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 and (1 ohms)I2 are negative because the traveling direction is the same direction of the current are negative
The term (10 ohms)I3 is positive because the traveling direction is the opposite direction of the current
Solve the summation to get the next equation
${\mathrm{l}}_2-2{\mathrm{l}}_3=1\mathrm{A}$

The10Vbatterysuppliesthe5Vbatteryandresistance109,sothecurrent(I1)?owsfrom10VbatterysplitsatpointctotwocurrentI2andI3(Seethe?gurebelow)-WeWillapplythejunctionruleinthiscasewherethejunctionruleisbasedon
conservation of electric charge and the current enters ajunction point is equal to the current flows out from this point, so in
our circuit, we could get the next as

$$\begin{gathered} {\mathrm{l}}_1={\mathrm{l}}_2+{\mathrm{l}}_3 \\ {\mathrm{l}}_1+{\mathrm{l}}_2=3\mathrm{A} \end{gathered}$$

NowWehavethreeequationswiththreevariables.LetusplugtheexpressionofI1fromequation(3)intoequation(I),hence

get a new form of equation (1) as next

$$\begin{gathered} 2+{\mathrm{l}}_3+{\mathrm{l}}_2=3\mathrm{A} \\ {\mathrm{l}}_2+{\mathrm{l}}_3/2=3/2\mathrm{A} \\ {\mathrm{l}}_2+{\mathrm{l}}_{3/}/2=3/2\mathrm{A} \\ + \\ -1{\mathrm{l}}_2-2{\mathrm{l}}_3=1\mathrm{A} \\ 5/2{\mathrm{l}}_3=1/2\mathrm{A} \\ {\mathrm{l}}_3=0.2\mathrm{A} \end{gathered}$$

ThisisthecurrentflOWsinthebottombranch-0.2A

The Current in the middle branch is

$$\begin{gathered} {\mathrm{l}}_2=1\mathrm{A}+2{\mathrm{l}}_3 \\ =1.4\mathrm{A} \end{gathered}$$

Therefrore the cureent in middle branch is 1.4A

The current in the upper branch is

$$\begin{gathered} {\mathrm{l}}_1={\mathrm{l}}_2+{\mathrm{l}}_3 \\ =1.6\mathrm{A} \end{gathered}$$

Therefore the current in the upper branch is 1.6A

Determine the potential differnece 

(b)Nowthepotentialdifferencebetweenthepointsaandb.Thepotentialbetweenaandbisthesumofthpotentialdropinthepathab.letusstartfroma.tobandtakethepathwiththeredcolorasshowninthefigurebelowwhere,inthispath,Wehavetworesist or

$$\begin{gathered} {\mathrm{V}}_{\mathrm{ab}}={\mathrm{V}}_{\mathrm{b}}-{\mathrm{V}}_{\mathrm{a}} \\ =3{\mathrm{l}}_1-4{\mathrm{l}}_2 \\ =-10.4\mathrm{V} \end{gathered}$$
Don't forget that the terms (3.00ohmsl) and (4.00 ohms) are negative because the traveling path that We take, is the samediirection of the current flows in both resistors as shOWn in the figure below. As the potential is negative, therefore, at point a
be potential is higher than at point b

Therefore the potential diffeernce at a is higher than b and equal to 10.4V