Calculate the current through each resistance in the given circuit. Also calculate the potential difference between the points a and b.
`E_1 = 6V, E_2 = 8V, E_3 =10V`,
`R_1 =5 Omega`,` R_2 =10 Omega, R_3 =4Omega`
Assume that all the cless have no internal resistance.

Explanation

The process of solving a circuit involves three steps:
i. Assume unkonwns (x, y, ….) for currents in different
branches of the circuit. Use Kirchhoff's Current law at
the junctins so that the number of unknowns introduced
is minimum. Let x be the current through `R_1` and y be the
current through `R_3` as shown in Fig. 5.74. Kirchhoff's
current law at the junction a gives a current (x-y)
through `R_2`

ii. Selectas many loops as the number of unknows introduced
for currents. Apply Kirchhoff's voltage law throgh every
loop. Going anticlockwise through the loop containing
`R_1 and R_3`(starting from junction a), we get
`+xR_1 - E_1 + yR_3 + E_3 = 0`
or 5x +4y = -4 ....(i)

Going clockwise through the loop containing `R_2 and R_3` (starting from juction a), we get
`-R_2(x-y) -E_2 +yR_3 + E_3 =0`
or 5x - 7y =1 (ii)
Some currents may come out to be negative. This simply
means that their directions were incorrectly assumed. so the signs of the currents will given us the correct direction
of each current. Solving Eqs. (i) and (ii) we get
`x = (-24)/(55) and y =(-5)/(11)A`
`:. x-y = (1)/(55)A`
This signs indicate that the direction of x and y were
assumed incorrectly, while the direction of (x-y) was
correct. So
current `i_1 ("through "R_1) = (24)/(55)`A toward left
current `i_2("through " R_2) = (1)/(55)` A toward right
current `i_3 ("through "R_3) =(5)/(11)` A upward
The current directions are shown in the circuit diagram

iii. Potential difference between a and b
The potentia difference between any two points in a
circuit is calculated by adding the changes in potential
while going through any path form one point to the other
point. Hence, let us go from b to a through `R_3.`
`V_a - V_b = +yR_3 + E_3 = ((-5)/(11))xx4+ 10 =(90)/(11)v`