These three basic rules/formulas are needed to solve the system of equations envolved:
To solve the equations we will need to name our components, we have chosen a system in which each component is named by a letter indicating it's type (I for current, E for the EMF, and R for resistance) and a subletter indicating it's branch (A, B, or C). Thus the current in the leftmost branch would be (by Ohm's Law): IA = VA/RA. Now we are ready to formulate the system.
First, we take the outer most loop ACDF and use the Junction
Rule. Let's take junction B, since we assume that all of the
currents flow from top to bottom, the sum of the currents heading out
of the loop is equal to zero (since no current is heading in) i.e.,
IA + IB + IC = 0.
Next we take loop ABEF and apply the Loop Rule, which says
that the sum of the potential rises must equal to the sum of the
potential drops i.e.,
EA + IA*RA = EB + IB*RB;
we also apply this rule to the outer loop, ACDF, getting:
EA + IA*RA = EC + IC*RC.
Let us summarize our system of three equations:
IA + IB + IC = 0;
EA + IA*RA = EC + IC*RC;
EA +IA*RA = EB + IB*RB;
Solving for the currents we get:
IB = (EA - EB )/RB + IA*(RA/RB);
IC = (EA - EC)/RC + IA*(RA/RC);
IA = ((EB - EA)/RB + (EC - EA)/RC) / (1 + (RA/RB ) + (RA/RC));
Therefore, since the resistance, R, and the EMF, E, are known for each of the branches, the currents, I, are calculated by the above method.
The voltage across the resistors is arrived at as follows: (let
V be the voltage across the resistor, by Ohm's Law)
V = IR;
Then we substitute any of the above values for I.
Example, VB = IB*RB = RB*((EA - EB)/RB + IA*(RA/RB)).