Question

In Fig. 1.86 compute the relative potentials of points A, B, C, D and E which (i) point A is grounded and (ii) point D is grounded. Does it affect the circuit operation or potential difference between any pair of points?

Answer 1

As seen, the two batteries have been connected in series opposition. 

Hence, net circuit voltage = 34 − 10 = 24 V 

Total circuit resistance = 6 + 4 + 2 = 12 Ω 

Hence, the circuit current = 24/12 = 2 A 

Drop across 2 Ω resistor = 2 × 2 = 4 V, 

Drop across 4 Ω resistor = 2 × 4 = 8 V 

Drop across 6 Ω resistor = 2 × 6 = 12 V

(i) Since point B is directly connected to the positive terminal of the battery whose negative terminal is earthed, hence V_B = + 34 V. 

Since there is a fall of 4 V across 2 Ω resistor, V_C = 34 − 4 = 30V 

As we go from point C to D i.e. from positive terminal of 10-V battery to its negative terminal, there is a decrease in potential of 10 V. Hence, V_D = 30 − 10 = 20 i.e. point D is 20 V above the ground A. 

Similarly, V_E = V_D − voltage fall across 4Ω resistor = 20 − 8 = + 12V

Also V_A = V_E − fall across 6Ω resistor = 12(2 × 6) = 0 V 

(ii) In Fig. 1.87, point D has been taken as the ground. Starting from point D, as we go to E there is a fall of 8 V. Hence, V_E = − 8 V. Similarly, V_A = − (8 + 12) = − 20 V.

As we go from A to B, there is a sudden increase of 34 V because we are going from negative terminal of the battery to its positive terminal. 

∴ V_B = − 20 + 34 = + 14V 

V_C = V_B − voltage fall across 2Ω resistor = 14 − 4 = + 10 V. 

It should be so because C is connected directly to the positive terminal of the 10V battery. 

Choice of a reference point does not in any way affect the operation of a circuit. Moreover, it also does not change the voltage across any resistor or between any pair of points (as shown below) because the ground current i_g = 0. 

Reference Point A 

V_CA = V_C − V_A = 30 − 0 = + 30 V; V_CE = V_C − V_E = 30 − 12 = + 18 V 

V_BD = V_B − V_D = 34 − 20 = + 14 V 

Reference Point D 

V_CA = V_C − V_A = 10 − (− 20) = + 30 V; 

V_CE = V_C − V_E = 10 − (− 8) = + 18 V 

V_BD = V_B −V_D = 14 − 0 = + 14 V