Find the potential difference between A & B in the given circuit.

$$\begin{gathered} Diode is in forward bias and behaves like conducting wire. \\ {R}_{eq} = 10+\frac{10\times 10}{10+10}= 15 k\Omega \end{gathered}$$

$$\begin{gathered} Ohm\text{'}s Law :\hspace{0.17em}Current pas\sin g through battery is i \\ i=\frac{V}{R}=\frac{30}{10+\frac{10\times 10}{10+10}}=2mA \end{gathered}$$

$$\begin{gathered} Now, current will divide in two equal parts. \\ {i}_{AB}=1mA \end{gathered}$$

$$\begin{gathered} \therefore Potential drop across AB \\ {V}_{AB}=i_{AB}{R}_{AB}={10}^{-3}\times 10\times {10}^3=10V \end{gathered}$$