Electronics - Diodes and Applications - Discussion
Discussion Forum : Diodes and Applications - General Questions (Q.No. 39)
39.
What is wrong with this circuit?
Discussion:
13 comments Page 1 of 2.
Mahesh Babu said:
4 years ago
I think the right answer is C.
(3)
Sravani said:
5 years ago
If the diode is close then the voltage across the diode is Vcc-Vd = 9v.
Syam said:
5 years ago
In case diode is close, then what is the voltage in voltmeter.
Bhuvaneswari said:
7 years ago
Thank you for your explanation @Shalini.
Virat said:
9 years ago
If a zener is open then the current is not open so the answer is 2.
Ankita said:
1 decade ago
Here the breakdown voltage of diode is 15V so, it may get conducted by the source of 24V. So, if it is in conduction then the reading of voltmeter(4.79V) will become wrong. So, To get that amount of voltage we have To keep diode open.
(1)
Tulsi said:
1 decade ago
Here the zener diode is reverse biased, and in reverse bias condition the resistance across it is very high, for this reason diode should be open.
Nag said:
1 decade ago
Parallel= voltage is same current different.
Series =voltage is different, current same.
But here r1 and r2 are parallel voltages are different .
So open the diode,r1 and r2 are series so different voltages
i=v/r; r=r1+r2;
i=24/12.5
i=1.92mA
v1=1.92mA*10k ohm=19.2v
v2=1.92mA*2.5K ohm=4.608v (4.7v)
v=v1+v2 = 19.2v+4.608v
=23.8v (24v)
Series =voltage is different, current same.
But here r1 and r2 are parallel voltages are different .
So open the diode,r1 and r2 are series so different voltages
i=v/r; r=r1+r2;
i=24/12.5
i=1.92mA
v1=1.92mA*10k ohm=19.2v
v2=1.92mA*2.5K ohm=4.608v (4.7v)
v=v1+v2 = 19.2v+4.608v
=23.8v (24v)
Chaitali said:
1 decade ago
Thanks shalini :-)
Vamshikrishna said:
1 decade ago
Good explanation, shalini.
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