[Solved] For the BJT, β=∞, VBEon=0.7V VCEsat=0.7V. The switch is initially closed. At t=0, it is opened. At which time the BJT leaves the active region?

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Q.	For the BJT, β=∞, VBEon=0.7V VCEsat=0.7V. The switch is initially closed. At t=0, it is opened. At which time the BJT leaves the active region?
A.	20ms
B.	50ms
C.	60ms
D.	70ms
Answer» B. 50ms
Explanation: at t < 0, the bjt is off in cut off region. ib=0 as β=∞, so ic=ie. when t > 0, switch opens and bjt is on. the voltage across capacitor increases. from the input loop, -5-vbe-i(4.3k)+10=0 and gives i=1ma. ic1=1-0.5=0.5ma. vc1=0.7+4.3+10=-5v. ic1=c1dvc1/dt. from

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