If the emitter resistor is open, the collector voltage is

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Examoogle · Accepted Answer

High

Explanation: When the emitter resistor is open-circuited, the emitter current $I_E$ drops to zero. Since the collector current $I_C$ depends on the emitter current ($I_C \approx I_E$), the collector current also becomes zero, causing the collector voltage $V_C$ to rise to the supply voltage $V_{CC}$ as there is no voltage drop across the collector resistor.

Working Principle: In a common-emitter configuration, the collector voltage is given by $V_C = V_{CC} - I_C R_C$. If the emitter resistor is open, the transistor is cut off ($I_C = 0$). Substituting $I_C = 0$ into the equation yields $V_C = V_{CC}$. Since $V_{CC}$ is typically higher than the quiescent collector voltage, the collector potential rises to the supply rail level, which is considered 'High'.

Key Formulas: $V_C = V_{CC} - I_C R_C$ — Collector voltage equation; $I_C \approx I_E$ — Relation between collector and emitter current

Key Points: An open emitter circuit forces the transistor into the cut-off state.; Without emitter current, no collector current flows through the load resistor $R_C$.; The absence of voltage drop across $R_C$ means the collector is pulled up to the supply potential $V_{CC}$.; This is a common diagnostic indicator for BJT circuit troubleshooting.

Applications: Fault diagnosis in BJT amplifier circuits; Transistor bias stabilization analysis

Additional Info: In a correctly biased BJT, $V_C$ is usually significantly lower than $V_{CC}$ due to the voltage drop $I_C R_C$.. Option A is incorrect because the collector is not grounded; Option C is incorrect as the current path is interrupted; Option D is incorrect as the state is deterministic.

Answer: B is correct — If the emitter resistor is open, the transistor enters the cutoff region, causing the collector voltage to rise to the supply voltage $V_{CC}$.

Exam Tip: When analyzing open-circuit faults in BJT circuits, always identify the component that defines the current flow ($I_E$ or $I_B$). If the path for these currents is broken, the transistor is driven to cut-off, and the output node will typically rise to the supply voltage.

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