A circuit consists of two batteries (with negligible internal resistance), five ohmic resistors (Figure 19.88). The connecting wires that have negligible resistance. The letters A through are shown to make it possible to refer to specific parts of the circuit.

(a) Write all the equations necessary to solve for the unknown currents ${\mathbf{I}}_{\mathbf{1}}$, ${\mathbf{I}}_{\mathbf{2}}$, ${\mathbf{I}}_{\mathbf{3}}$, ${\mathbf{I}}_{\mathbf{4}}$ and ${\mathbf{I}}_{\mathbf{5}}$, whose directions are indicated on the circuit diagram. Do not solve the equations but do explain very clearly what your equations are based on and to what they refer.

Assume that a computer program has solved your equations in terms of known battery voltages and known resistances so that the currents ${\mathbf{I}}_{\mathbf{1}}$, ${\mathbf{I}}_{\mathbf{2}}$,${\mathbf{I}}_{\mathbf{3}}$ ,${\mathbf{I}}_{\mathbf{4}}$and ${\mathbf{I}}_{\mathbf{5}}$are are known. (b) In terms of known quantities calculate ${\mathbf{V}}_{\mathbf{D}}\mathbf{-}{\mathbf{V}}_{\mathbf{A}}$and check that your sign makes sense. (c) In terms of known quantities, calculate the power produced in battery number 2.

The potential of the battery 1 is $V_1$

The potential of the battery 2 is$V_2$

The resistance of the first resistor is $R_1$

The resistance of the second resistor is $R_2$

The resistance of the third resistor is $R_3$

The resistance of the fourth resistor is $R_4$

The resistance of the fifth resistor is $R_5$

The current through first resistor is $I_1$

The current through second resistor is $I_2$

The current through third resistor is $I_3$

The current through fourth resistor is $I_4$

The current through fifth resistor is $I_5$

The conservation of charge at every node and conservation of potentials in every loop is used to solve the above problem. The conservation of charge says that the incoming currents at a node are equal to the outgoing current from that node. The conservation of potential says that the net potential in any loop is always zero

Apply the Kirchoff’s current law at C.

$I_1+I_2=I_3+I_4$

The current $I_1$ and $I_2$ are incoming current at node C and $I_3$, $I_4$ are outgoing current at node C.

Apply the Kirchhoff’s current law at F.

$I_3=I_2+I_5$

The current $I_3$ is incoming current at node F and $I_2$, $I_5$ are outgoing current at node F.

Apply the Kirchoff’s voltage law in the loop 1.

$V_1-I_1{R}_1-I_4{R}_4=0$

The direction of current in loop 1 is clockwise so the sign of the current in loop 1 is negative.

Apply the Kirchoff’s voltage law in the loop 2.

$-V_2+I_2{R}_2+I_3{R}_3=0$

The direction of current in loop 2 is anti-clockwise so the sign of the current in loop 1 is positive.

Apply the Kirchoff’s voltage law in the loop 3.

$-I_4{R}_4+I_3{R}_3+I_5{R}_5=0$

The clockwise current in loop 3 is positive while the anti-clockwise current in loop 3 is negative.

Therefore, the set of equations for unknown currents $I_1+I_2=I_3+I_4$, $I_3=I_2+I_5$, $V_1-I_1{R}_1-I_4{R}_4=0$, $-V_2+I_2{R}_2+I_3{R}_3=0$and $-I_4{R}_4+I_3{R}_3+I_5{R}_5=0$.