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A silicon transistor with VBE sat = 0.8 V, b dc = 100 and VCE sat = 0.2 V is used in the circuit shown in given figure What is the minimum value of RC for which transistor is in saturation ?
  • a)
    4286W
  • b)
    4667W
  • c)
    500W
  • d)
    1000W
Correct answer is 'B'. Can you explain this answer?
Verified Answer
A silicon transistor with VBE sat = 0.8 V, b dc = 100 and VCE sat = 0....
 
Step 1
Apply kvl in 1 loop
5 - IbRb - 0.8 =0
5 - Ib*(200*1000ohm) - 0.8 =0
Base current (Ib)=(5-0.8)/(200*1000ohm)
=(4.2)/(200*1000)
=(2.1)/(100000)

2nd step
we know, collector current (Ic)= beta times of base current
so, Ic=(100*2.1)/(100000)=0.0021
 
3 step:-
Apply kvl in 2 loop
10 - 0.0021*Rc - 0.2 =0
minimum value of Rc is
Rc=(9.8)/(0.0021)
=4666.66ohm
=4667 ohm
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Most Upvoted Answer
A silicon transistor with VBE sat = 0.8 V, b dc = 100 and VCE sat = 0....
1 step:-Apply kvl in 1 loop
5 - IbRb - 0.8 =0
5 - Ib*(200*1000ohm) - 0.8 =0
Base current (Ib)=(5-0.8)/(200*1000ohm)
=(4.2)/(200*1000)
=(2.1)/(100000)
2 step:- we know, collector current (Ic)= beta times of base current
so, Ic=(100*2.1)/(100000)=0.0021
3 step:- Apply kvl in 2 loop
10 - 0.0021*Rc - 0.2 =0
minimum value of Rc is
Rc=(9.8)/(0.0021)
=4666.66ohm
=4667 ohm
Free Test
Community Answer
A silicon transistor with VBE sat = 0.8 V, b dc = 100 and VCE sat = 0....
 
Step 1
Apply kvl in 1 loop
5 - IbRb - 0.8 =0
5 - Ib*(200*1000ohm) - 0.8 =0
Base current (Ib)=(5-0.8)/(200*1000ohm)
=(4.2)/(200*1000)
=(2.1)/(100000)

2nd step
we know, collector current (Ic)= beta times of base current
so, Ic=(100*2.1)/(100000)=0.0021
 
3 step:-
Apply kvl in 2 loop
10 - 0.0021*Rc - 0.2 =0
minimum value of Rc is
Rc=(9.8)/(0.0021)
=4666.66ohm
=4667 ohm
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A silicon transistor with VBE sat = 0.8 V, b dc = 100 and VCE sat = 0.2 V is used in the circuit shown in given figure What is the minimum value of RC for which transistor is in saturation ?a)4286Wb)4667Wc)500Wd)1000WCorrect answer is 'B'. Can you explain this answer?
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