Electronic – Voltage sources in parallel Kirchhoff volage law

kirchhoffs-lawsvoltage

Can somebody please tell me how can I apply Kirchhoff voltage law in the following circuit??
Is there a way to find an equivalent circuit for the sources V1,V2 and resistors R1 and R2??

schematic

simulate this circuit – Schematic created using CircuitLab

Best Answer

We haven't yet covered the general case where R1 and R2 are different. You can use nodal analysis or a Thevenin equivalent, but you can also take a more mathematical approach and use plain KVL. Each path around the circuit gives a different equation. Let's say \$R_1 = 200\ \Omega\$ and \$R_2 = 500\ \Omega\$. We'll define \$I_1\$ as the current through \$R_1\$ and \$I_2\$ as the current through \$R_2\$. There are two possible paths:

$$5\ \mathrm V - 4700\ \Omega \cdot (I_1 + I_2) - 0.6\ \mathrm V - 200\ \Omega \cdot I_1 = 0$$

$$5\ \mathrm V - 4700\ \Omega \cdot (I_1 + I_2) - 0.6\ \mathrm V - 500\ \Omega \cdot I_2 = 0$$

That's two equations in two unknowns, which means we can find a solution. We can solve for the currents by hand or using a convenient tool such as Wolfram Alpha. The solutions are:

$$I_1 \approx 650\ \mathrm{\mu A}$$ $$I_2 \approx 260\ \mathrm{\mu A}$$

Now consider the case where the two resistors are equal \$(R_1 = R_2 = 270\ \Omega)\$.

$$5\ \mathrm V - 4700\ \Omega \cdot (I_1 + I_2) - 0.6\ \mathrm V - 270\ \Omega \cdot I_1 = 0$$ $$5\ \mathrm V - 4700\ \Omega \cdot (I_1 + I_2) - 0.6\ \mathrm V - 270\ \Omega \cdot I_2 = 0$$

We can solve these as simultaneous equations just like before. But notice how similar the equations are! Nothing changes except the name of the variable. But the name of the variable is meaningless -- we can number the currents however we want. Since we can swap the two equations (and thus the two branches) without being able to tell the difference, \$I_1\$ must equal \$I_2\$. We say that the two branches are symmetrical. Symmetry is a powerful concept in mathematics and the sciences. It allows us to simplify our analysis by eliminating variables. If you've studied Gauss's Law in physics, you've already seen symmetry in action.

All this is a fancy mathematical way of saying that when you have two identical branches in parallel, their currents must be the same. That lets us write a single equation:

$$5\ \mathrm V - 4700\ \Omega \cdot (I_{12} + I_{12}) - 0.6\ \mathrm V - 270\ \Omega \cdot I_{12} = 0$$

$$4.4\ \mathrm V - (2 \cdot 4700\ \Omega + 270\ \Omega)I_{12} = 0$$

$$I_{12} \approx 455\ \mathrm{\mu A}$$

where \$I_{12}\$ is the current through either one of the branches.

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