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(C) $p-n$ junction diode is forward biased when the potential at the $p$-side is higher than the potential at the $n$-side.$(A)$ $p$-side is at $+5 \

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$A, C$

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(C) $p-n$ junction diode is forward biased when the potential at the $p$-side is higher than the potential at the $n$-side.$(A)$ $p$-side is at $+5 \ V$ and $n$-side is at $+10 \ V$. Since $5 \ V $(B)$ $p$-side is at $-10 \ V$ and $n$-side is at $0 \ V$ (ground). Since $-10 \ V $(C)$ $p$-side is at $-12 \ V$ and $n$-side is at $-5 \ V$. Since $-12 \ V $(D)$ $p$-side is at $0 \ V$ (ground) and $n$-side is at $+5 \ V$. Since $0 \ V Wait,let's re-examine the diode symbols in the image:$(A)$ Diode points towards $+5 \ V$. $p$-side is at $+10 \ V$ (via resistor),$n$-side is at $+5 \ V$. $10 > 5$,so forward biased.$(B)$ Diode points towards ground. $p$-side is at $-10 \ V$,$n$-side is at $0 \ V$. $-10 $(C)$ Diode points towards $-12 \ V$. $p$-side is at $-5 \ V$,$n$-side is at $-12 \ V$. $-5 > -12$,so forward biased.$(D)$ Diode points towards ground. $p$-side is at $0 \ V$,$n$-side is at $+5 \ V$. $0 Thus,diodes $(A)$ and $(C)$ are forward biased.

In the following figure,the diodes which are forward biased,are

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