The potential difference between A and B in the following figure is-

 A. $32V$
 B. $48V$
 C. $24V$
 D. $14V$

Hint: You can start by briefly defining Ohm’s law and its equation $\dfrac{V}{I} = R$. Then use Kirchhoff's law (for any closed system the total voltage drop between any two points is equal to the sum of all the individual potential drops between the same two points) to calculate the potential drop between point A and B. Remember the potential drop across a resistor is $V = IR$ and the potential drop across a battery is equal to its e.m.f.

Complete step by step answer:
To solve this problem we have to first understand Ohm’s law and the equation obtained from it.
According to Ohm’s law, the potential difference between any two points is directly proportional to the current between those two points, i.e.
$V \propto I$
$\dfrac{V}{I} = R$
Here, $V = $ The potential difference between the two points
$I = $ Current
$R = $ Resistance
So, $V = IR$
This equation gives us the value of potential drop across a resistor or a conductor.
According to Kirchhoff’s law for any closed system, the total voltage drop between any two points is equal to the sum of all the individual potential drops between the same two points.
So, applying Kirchhoff’s law between points A and B, we get
${V_{AB}} = $ Potential drop across 6 ohm resistor + Potential drop across battery of 6 e.m.f + Potential drop across 9 ohm resistor + Potential drop across battery of 4 e.m.f. + Potential drop across 5 ohm resistor
${V_{AB}} = I\left( 6 \right) + 12 + I\left( 9 \right) - 4 + I\left( 5 \right)$
$ \Rightarrow {V_{AB}} = 2\left( 6 \right) + 12 + 2\left( 9 \right) - 4 + 2\left( 5 \right)$
$ \Rightarrow {V_{AB}} = 48V$
The potential difference between points A and B is $48V$ .
Hence, option B is the correct choice.

Note:
It is very important to remember that the value of the potential drop that we calculated across the battery in this solution is for an ideal battery. In reality, the battery would itself have some internal resistance, this would lead to further potential drop aside from the potential drop due to the e.m.f. of the battery.