Calculate the current through the given circuit (the diodes are ideal)

(A) 6A
(B) 12A
(C) 1As
(D) 2A

Hint: For solving the circuit diagram first we use the biasing of diode i.e. if the diode is reversed biased then there is no current flow in the circuit. If the diode is forward biased then there is current flow in the circuit. If resistance are in series then total resistance is ${R_1} + {R_2} + ....$. After then applying ohm’s law we will get the value of current flowing in the circuit.

Complete step by step answer:
In the circuit diagram, we can sec ${D_1}$ is reversed biased that means there are no current flow through ${D_1}$
${D_2}$ is forward biased that means there are current flow through ${D_2}$
So the total resistance of the circuit is
${R_L} + {R_{{D_2}}}$
$\Rightarrow{T_{total}} = 5 + 10$
$\Rightarrow{R_{total}} = 15\Omega $
We know according to ohm’s law current through the circuit
$I = \dfrac{v}{r}$
Given that $v = 30\,V$
${R_{total}} = 15\Omega $
So, $I = \dfrac{{30}}{{15}}$
$\therefore I = 2\,A$

Hence option D is correct.

Note: Forward biasing means p part of p-n junction diode is connected to $ + ve$ terminal of battery and $ - ve$ terminal to n junction. Reverse biasing means p terminal is connected to $ - ve$ terminal and n junction is connected to $ + ve$ terminal of battery