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(B) diode is forward biased when the potential at the $P$-terminal $(V_P)$ is higher than the potential at the $N$-terminal $(V_N)$.In option $A$: $V_

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(B) diode is forward biased when the potential at the $P$-terminal $(V_P)$ is higher than the potential at the $N$-terminal $(V_N)$.In option $A$: $V_P = +2 \ V$,$V_N = +3 \ V$. Since $V_P In option $B$: $V_P = +2 \ V$,$V_N = -2 \ V$. Since $V_P > V_N$,it is forward biased.In option $C$: $V_P = -2 \ V$,$V_N = +2 \ V$. Since $V_P In option $D$: $V_P = +2 \ V$,$V_N = +2 \ V$. Since $V_P = V_N$,there is no bias (or zero bias).Therefore,the correct option is $B$.

In the given options,the diode that is forward biased is

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