Question

In the following figure, what is the value of the potential at points a and b?

Answer 1

It depends on how the polarities of the two batteries are placed.
Here I assume that the battery on the left side has polarity- on top and polarity-+on bottom, while the battery on the right side has polarity- on bottom and polarity+ on top, so the current is flowing in anti-clockwise direction.

To solve the problem, we must first calculate the current flowing in the circuit. For this, we use Kirchoff's voltage law:
$\sum V_i =0$ (1)
We must consider both the emf of the batteries and the voltage drops on the resistors, which for sign convention we must take as negative. So (1) becomes
$15 V - (1 \Omega \cdot I)+5 V -(4 \Omega \cdot I)=0$
where I is the current flowing.
By solving the equation, we get:
$20 V=5 I$
$I= 4 A$
So, assuming the potential at top left corner is 0 V (because it is at same potential of the negative polarity of the 15 V battery), then the voltage drop on the $4 \Omega$ resistor is $\Delta V = IR=(4 A)(4 \Omega)=16 V$, and so the potential at the top right corner is 16 V. Instead, the bottom left corner is at same potential of the positive terminal of the 15 V battery, so at 15 V.

In case the polarities of the 5 V battery are inverted, the exercise must be re-done in the same way by changing the sign of the emf of the 5 V battery in the Kirchoff's law.