Kirchhoffs Law – For the circuit shown in

Solution :

Consider the two battery circuits of Figure given separately. Current in the 20 V battery circuit ABCD is 20 (6 + 5 + 9) = 1A. Similarly, current in the 40 V battery circuit EFGH is = 40/(5 + 8 + 7) = 2A. Voltage drops over different resistors can be found by using Ohm’s law. For finding VCE i.e. voltage of point C with respect to point E, we will start from point E and go to C via points H and B. We will find the algebraic sum of the voltage drops met on the way from point E to C. Sign convention of the voltage drops and battery e.m.fs. would be the same as discussed in Article Determination of Voltage Sign.

∴ V_CE = (− 5 × 2) + (10) − (5 × 1) = − 5V

The negative sign shows that point C is negative with respect to point E.

V_AG = (7 × 2) + (10) + (6 × 1) = 30 V.

The positive sign shows that point A is at a positive potential of 30 V with respect to point G.