Question

In the circuit shown in Fig. E26.28, find (a) the current in each branch and (b) the potential difference Vabof point a relative to point b.

Short answer

The current in upper branch is 0.8A , the current in middle branch is 0.2A ,the current in lower branch is 0.6A

The potential differnece is 3.2V and the potential at a is lower than b

Step-by-step solution

Step 1:About potential difference

The difference in potential between two points that represents the work involved or the energy released in the transfer of a unit quantity of electricity from one point to the other

Determine te current in each branch

a) the current in each branch- the next; the current in the top branch is (right to left ), in themiddle branch is (left to right) and in the bottom branch is (left to right)- So our target is 1, 2 and 3- We will
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 and the algebraic sum ofthese differences is zero when We return to the starting point the loop directions and the paths the We Will take to get the target variables.
let us use loop 1 (Closed blue path) and apply equation 26-6 as ShOWn in the direction of our travel
counterclockwise

$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ -3{\mathrm{l}}_1+10\mathrm{V}-2{\mathrm{l}}_1-1{\mathrm{l}}_{2}4{\mathrm{l}}_2-5\mathrm{V}=0 \end{gathered}$$

10,Vispositivebecausethedirectionoftravelingisfromnegativetopositiveterminalinthebattery(Seefigure26.8a)-
The terms (3 Ohms)I1, (2 (2)11, (1 ohms)I2 and (4 ohms)I2 are negative because the

traveling direction is in the direction of thecurrent (See figure 26.8b )-
5,V is negative because the direction of traveling is from positive to negative terminal in the battery
We can solve the summation to get the next equation

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

the same steps let use loop 2 (Closed black path) and apply equation 26.6 as shown in the figgure below where the

direction of our travel is counterclockwise

$$\begin{gathered} \sum _{}\mathrm{V}=0 \\ 4{\mathrm{l}}_2+5\mathrm{V}+1{\mathrm{l}}_2-10{\mathrm{l}}_3=0 \end{gathered}$$

Vispositivebecausethedirectionoftravelingisfromnegativetopositiveterminalinthebattery(See?gure26.8a)-
The termsare positive because the traveling direction is in the opposite direction of the c
The term (10 ohms)I3 is negative because the traveling direction is the same direction of the current . Now

${\mathrm{l}}_2-2{\mathrm{l}}_3=-1\mathrm{A}$

The10Vbatterysuppliesthe5Vbatteryandresistance10,sothecurrent(I1)
two current I2 and I3 Apply the junction rule in this case. Where the junction rule is based or
conservation of electric charge and the current enters ajunction point is equal to the current flows out from this point,
our circuit, we could get the next as I1 enters (1 while I2 and I3 ?ow out from (1
$$\begin{gathered} {\mathrm{l}}_1+{\mathrm{l}}_2+{\mathrm{l}}_3 \\ {\mathrm{i}}_1+{\mathrm{l}}_2=1\mathrm{A} \end{gathered}$$

Nowwehavethreeequationswiththreevariables-LetusplugtheexpressionofI1fromequation(3)intoequation(1),henc
we will get a new form of equation (1) as next

role="math" localid="1664208993557" $$\begin{gathered} {\mathrm{l}}_1+{\mathrm{l}}_2+{\mathrm{l}}_3 \\ {\mathrm{i}}_1+{\mathrm{l}}_2=1\mathrm{A} \\ {\mathrm{l}}_2+{\mathrm{l}}_3+{\mathrm{l}}_2=1\mathrm{A} \\ 2{\mathrm{l}}_2+{\mathrm{l}}_3=1\mathrm{A} \\ {\mathrm{l}}_2+{\mathrm{l}}_3/2=\frac{1}{2}\mathrm{A} \end{gathered}$$

Nowwecanmultiplyequation(2)by—1andaddittoequation(1*)togetthenext
$$\begin{gathered} {\mathrm{l}}_2+{\mathrm{l}}_3/2=1/2\mathrm{A}+-1\left({\mathrm{l}}_2-2{\mathrm{l}}_3=-1\mathrm{A}\right) \\ 5/2{\mathrm{l}}_3=3/2\mathrm{A} \\ {\mathrm{l}}_3=0.6\mathrm{A} \end{gathered}$$
This is the current flows in the lower branch-

Therefore the current in upper branch is 0.8A , the current in middle branch is 0.2A ,the current in lower branch is 0.6A

Step 3:Determine the Current flow inj upper and lower branch  

NowWecanplugthevalueforI3intoequation(2)togetI2asnext
$$\begin{gathered} {\mathrm{l}}_2=-1\mathrm{A}+2{\mathrm{l}}_3 \\ =-1\mathrm{A}+2.0.6\mathrm{A} \\ =0.2\mathrm{A} \end{gathered}$$
Therefore is the current flows in the center branch-0.2A

Again,letusplugthevaluesforI2andI3intoequation(3)togetI1

$$\begin{gathered} 1={\mathrm{l}}_2+{\mathrm{l}}_3 \\ =0.2\mathrm{A}+0.6\mathrm{A} \\ =0.8\mathrm{A} \end{gathered}$$
Therefore is the current flow in the upper branch-0.8A

Determine the potential differnce

(b)NowWewantto?ndthepotentialdifferencebetweenthepointsa.andb-Thepotentialbetweena.andbisthesumoft
potential drop in the path let

startfromatobandtakethepathwiththeredcolorasshOWninthefigurebelow
where, in this path,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{ab}}={\mathrm{V}}_{\mathrm{b}}-{\mathrm{V}}_{\mathrm{a}} \\ =3{\mathrm{l}}_1+4{\mathrm{l}}_2 \\ =3.2\mathrm{V} \end{gathered}$$

Don't forget that the terms are positive because the travel ina path that we take. is in theopposite direction of the current flOWS in both resistors as shown in the figure belo w. As the potential is positive, therefore, at point a the potential is lower than at point b

Therefore the potential differnece is 3.2V and the potential at a is lower than b