question_answer

                        \[I=\frac{-M}{R}\frac{dp}{dl}\]

(i) In the circuit given below, is the diode forward biased or reverse biased?

(ii) What is the current flowing in the circuit given below?

Answer:

(i) The diode is forward biased as the p-side of the diode is connected to zero potential (ground) and n-side of the diode is connected to-10V. Therefore, p-side is at a higher potential than the n-side.

(ii) Here \[{{D}_{1}}\] is reverse biased and \[{{D}_{2}}\] is forward biased. So the circuit can be redrawn as

Net resistance of the circuit, \[R=4\Omega +2\Omega =6\Omega \] Voltage = 12V \[\therefore \] The current flowing, \[I=\frac{V}{R}=\frac{12}{6}=2A\]