Question

in cutoff, vce is
select one:
o a. approximately 0.2 v
o b. equal to vcc
o c 0.7 v
o d. undefined

Explanation:

In the cutoff region of a BJT (Bipolar Junction Transistor), the collector - emitter voltage (\(V_{CE}\)) is approximately equal to the supply voltage \(V_{CC}\). This is because when the transistor is in cutoff, the collector current \(I_C\) is approximately zero (neglecting leakage current). Using Kirchhoff's voltage law in the collector - emitter loop (\(V_{CC}=V_{CE}+I_CR_C\)), when \(I_C\approx0\), \(V_{CE}\approx V_{CC}\). Option a (0.2 V) is the approximate \(V_{CE}\) in the saturation region, option c (0.7 V) is the approximate base - emitter voltage (\(V_{BE}\)) for a silicon transistor in the active region, and option d is incorrect as \(V_{CE}\) is well - defined in cutoff.

Answer:

b. equal to \(V_{CC}\)