• # 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?

 (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, $I=\frac{V}{R}=\frac{12}{6}=2A$