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(A) $1$. When the multimeter is connected between $P$ and $Q$,no conduction is observed. This implies that $P$ and $Q$ are of the same type (both $n$-

Vedclass · Accepted Answer

It is a $PNP$ transistor with $R$ as collector

Vedclass · Answer

(A) $1$. When the multimeter is connected between $P$ and $Q$,no conduction is observed. This implies that $P$ and $Q$ are of the same type (both $n$-type or both $p$-type). Since the transistor is working,$P$ and $Q$ must be the collector and emitter,and $R$ must be the base.$2$. When the negative terminal of the multimeter is connected to $R$ (base) and the positive terminal is connected to $P$ or $Q$ (collector/emitter),conduction is observed. This means the base-collector and base-emitter junctions are forward-biased when the base is at a lower potential than the collector/emitter.$3$. In an $NPN$ transistor,the base is $p$-type and the collector/emitter are $n$-type. Forward biasing occurs when the $p$-type base is connected to the positive terminal and $n$-type regions to the negative terminal. However,the problem states the negative terminal is at $R$. This indicates the base is $n$-type and the collector/emitter are $p$-type,which corresponds to a $PNP$ transistor. $4$. Re-evaluating the standard multimeter test: If the negative terminal is at $R$ and conduction occurs,$R$ must be $p$-type. Thus,$R$ is the base of a $PNP$ transistor.

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