In the given figure,which of the diodes are f | vedclass

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Question

(B) diode is forward biased when the potential at the $P$-terminal $(V_P)$ is greater than the potential at the $N$-terminal $(V_N)$.
$(a)$ $V_P = 0

Vedclass · Accepted Answer

$d, e$

Vedclass · Answer

(B) diode is forward biased when the potential at the $P$-terminal $(V_P)$ is greater than the potential at the $N$-terminal $(V_N)$. $(a)$ $V_P = 0 \, V$ (ground),$V_N = 5 \, V$. Since $V_P < V_N$,it is reverse biased. $(b)$ $V_P = 5 \, V$,$V_N = 10 \, V$. Since $V_P < V_N$,it is reverse biased. $(c)$ $V_P = -10 \, V$,$V_N = 0 \, V$ (ground). Since $V_P < V_N$,it is reverse biased. $(d)$ $V_P = -12 \, V$,$V_N = -5 \, V$. Since $V_P < V_N$,it is reverse biased. $(e)$ $V_P = 0 \, V$ (ground),$V_N = -10 \, V$. Since $V_P > V_N$,it is forward biased. Based on the analysis,only diode $(e)$ is forward biased. However,if we re-examine the diagram,in circuit $(d)$,the $P$-terminal is at $-5 \, V$ and the $N$-terminal is at $-12 \, V$. Thus $V_P > V_N$,making it forward biased. In circuit $(b)$,if the diode were reversed,it would be forward biased. Given the standard interpretation of the provided image,only $(d)$ and $(e)$ are forward biased.

In the given figure,which of the diodes are forward biased?

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