For the circuit shown in Fig. 2.10, find VCE and VAG.

For the circuit shown in Fig. 2.10, find VCE and VAG.

 Solution. Consider the two battery circuits of Fig. 2.10 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 Art. 2.3. 

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

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

VAG = (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.