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Calculate the potentials of point A, B, C and D in Fig.What would be the new potential values if connections of 6-V battery are reversed? All resistances are in ohm.

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Since the two batteries are connected in additive series, total voltage around the circuit is = 12 + 6 = 18V. The drops across the three resistors as found by the voltage divider rule as shown in Fig. (a) which also indicates their proper polarities. The potential of any point in the circuit can be found by starting from the ground point G (assumed to be at 0V) and going to the point either in clockwise direction or counter-clockwise direction. While going around the circuit, the rise in potential would be taken as positive and the fall in potential as negative. (Art. 2.3). Suppose we start from point G and proceed in the clockwise direction to point A. The only potential met on the way is the battery voltage which is taken as positive because there is a rise of potential since we are going from its negative to positive terminal. Hence, VA is + 12V. 

VB = 12 − 3 = 9V; VC = 12 − 3 − 6 = 3V

Similarly, VD = 12 − 3 − 6 − 9 = − 6V. 

It is also obvious that point D must be at − 6V because it is directly connected to the negative terminal of the 6-V battery. 

We would also find the potentials of various points by starting from point G and going in the counter-clockwise direction. For example, VB = − 6 + 9 + 6 = 9 V as before. 

The connections of the 6 − V battery have been reversed in Fig. 1.92 (b).

 

Now, the net voltage around the circuit is 12 − 6 = 6 V. The drop over the 1 Ω resistor is = 6 × 1/(1 + 2 + 3) = 1 V; 

Drop over 2 Ω resistor is = 6 × 2/6 = 2 V. 

Obviously, VA = + 12 V, VB = 12 − 1 = 11 V, VC = 12 − 1 − 2 = 9 V. 

Similarly, VD = 12 − 1 − 2 − 3 = + 6V.

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