Electronic Devices - Transistor Bias Circuits - Discussion
Discussion Forum : Transistor Bias Circuits - General Questions (Q.No. 25)
25.
Refer to this figure. The value of IC is
10 
A.
A.10 mA.
5 mA.
50 mA.
Discussion:
4 comments Page 1 of 1.
Right?? said:
1 decade ago
We can't apply the law V=IR, directly for all case. We have to find the loop equation first. If VC=8V then, Ic = (Vcc-Vc) /RC = (12-8) /1.2K=3.3mA is the right answer.
But if we apply V = IR, Ic = Vc/RC = 8/1.2K = 6.6mA.
But if we apply V = IR, Ic = Vc/RC = 8/1.2K = 6.6mA.
Sanju prajapati said:
1 decade ago
Vce = Vc = 6v.
Vcc = Ic*Rc+Vce or Vcc = Ic*Rc+Vc.
(Vcc-Vc)/Ic = Rc.
(12-6)/1.2*10^(-3) = 5*10^(-3).
= 5mA.
Vcc = Ic*Rc+Vce or Vcc = Ic*Rc+Vc.
(Vcc-Vc)/Ic = Rc.
(12-6)/1.2*10^(-3) = 5*10^(-3).
= 5mA.
M.SENTHIL said:
1 decade ago
From figure,
Vc = 6V.
Rc = 1.2K.
Apply ohm law V = IR.
Ic = Vc/Rc.
Ic = 6/1.2 = 5mA.
Vc = 6V.
Rc = 1.2K.
Apply ohm law V = IR.
Ic = Vc/Rc.
Ic = 6/1.2 = 5mA.
Praveen sagar said:
1 decade ago
From Figure
Vcc=IcRc+Vc
Ic=Vcc-Vc/Rc
Ic=5mA
Vcc=IcRc+Vc
Ic=Vcc-Vc/Rc
Ic=5mA
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