Find the potential difference between the points A and B, ${V_{AB}}$:

A) $10V$
B) $20V$
C) $30V$
D) $None$

Hint: In order to solve this question, the concept of forward and reverse bias of the diode is important. See the circuit carefully and find out whether the diode is connected in forward bias or reverse bias. Also, carefully calculate the resistance across the branch $AB$ so that the potential difference can be calculated carefully.

Complete step by step solution:
Here in this circuit the diode shown is connected in forward bias.
As we can see that the positive terminal of the given battery of $30V$ is connected to the positive terminal of the diode shown in the given circuit. So the given circuit diagram in the question can be redrawn as the diagram shown below:

Here in the above circuit diagram the resistor $AB$ of $10\Omega $ is in parallel combination with an another resistor of $10\Omega $ so their combined resistance would be,
$\dfrac{1}{{{R'}}} = \dfrac{1}{{10}} + \dfrac{1}{{10}} = \dfrac{2}{{10}}$
On simplifying we get,
${R'} = \dfrac{{10}}{2} = 5\Omega $
Now the combined resistance of the circuit is given by,
$R = 10\Omega + 5\Omega $
On simplifying we get,
$R = 15\Omega $
Here the current flowing through the circuit is given by the expression,
$I = \dfrac{V}{R}$
Now putting the values of $V$ and $R$ . We have,
$I = \dfrac{{30V}}{{15\Omega }}$
Now the potential difference across $AB$ is given by,
${V_{AB}} = I{R_{AB}}$
Putting the values we have,
${V_{AB}} = \dfrac{{30V}}{{15\Omega }} \times 5\Omega $
On solving we have,
${V_{AB}} = 10V$

So, the potential drop across $AB$ is $10V$.

Note: Forward biasing is the condition in which the external voltage is delivered across the P-N junction diode. In the condition of forward bias setup, the P-side of the diode is connected to the positive terminal and N-side is attached to the negative side of the battery.