Figure 1of 1 Figure 1 B of 1 2 112 SolutionThe batteries are

Solution

The batteries are trying to push the current in opposite directions.

Overall effect is a single battery of emf = 9 - 6 = 3 V

Total resistance = 2 + 1 = 3 ohms

Current I = 3 / 3 = 1.0 A

The voltage drop across the 2 ohm resistor = V = I R = 1 x 2 = 2 V

The voltage drop across the 1 ohm resistor is = V = I R = 1 x 1 = 1 V

So, the drop across the first resistor is delta V = I R₁ = 2 V

The potential at point a = 9 V

The potential at point b is = 9 - 2 = 7 V

The battery connected backwards fights the 7 V, dropping it to 7 - 6 = 1 V.

So, the potential at point c = 1 V.

Last resistor which drops the potential by delta V = I R₂ = 1 V

net potential = 1 - 1 = 0 V at point d