Electronics - Diodes and Applications - Discussion
Discussion Forum : Diodes and Applications - General Questions (Q.No. 16)
16.
What is the current through the diode?
Discussion:
26 comments Page 2 of 3.
Karthik N said:
1 decade ago
I=?
I=v/r v=12 volt, r=12 k ohms or 1200 ohms
I=12/1200
I=0.01amps or 10 milliamps
I=v/r v=12 volt, r=12 k ohms or 1200 ohms
I=12/1200
I=0.01amps or 10 milliamps
Remya raghavan said:
1 decade ago
I=V/R.
V=12v.
R=12kohm.
I=12/12000.
I=0.001 amp.
I =1mA.
V=12v.
R=12kohm.
I=12/12000.
I=0.001 amp.
I =1mA.
SHAHNWAZ said:
1 decade ago
It is right in case of ideal vd=0 because internal diod resistance is 0.
So I=12-0/12=1mA.
So I=12-0/12=1mA.
Ayush sharma said:
1 decade ago
In case of ideal diode in forward bias voltage drop is 0.
Hence I= V/R.
I=12/12k.
=1mA.
Hence I= V/R.
I=12/12k.
=1mA.
Kishor A patel said:
1 decade ago
Generally si or ge diode has consider 0.7v barrier. But in case of ideal it should be consider 0v, that's why answer is 1mA.
Ramdas tibile said:
1 decade ago
In ideal condition drop is equal to 0 so answer is 1 mA.
Tabrej said:
1 decade ago
Since diode is ideal, voltage drop across diode is zero. So current through diode i=12/12k=1ma. Nothing else.
Mohammed Arif Rana said:
1 decade ago
We will assume diode as ideal because there is no specify that it is Si or Ge.
So I = v/r, given v=12v & r=12K.
I = 12v/12k = 1mA.
So I = v/r, given v=12v & r=12K.
I = 12v/12k = 1mA.
Imran said:
9 years ago
As they haven't mentioned about diodes condition whether it is ideal or not.
so,assuming;
1) diode as ideal = drop will be 0v,
Hence answer = 1mA.
2) diode as practical(non-ideal) = drop will be 0.7 for silicon diode.
Hence answer = 0.942mA.
so,assuming;
1) diode as ideal = drop will be 0v,
Hence answer = 1mA.
2) diode as practical(non-ideal) = drop will be 0.7 for silicon diode.
Hence answer = 0.942mA.
Himanshu said:
9 years ago
V = IR.
12 = I * 12000.
I = 12/12000.
I = 1mA.
IN IDEAL DIODE, THERE IS NO VOLTAGE DROP DOWN.
12 = I * 12000.
I = 12/12000.
I = 1mA.
IN IDEAL DIODE, THERE IS NO VOLTAGE DROP DOWN.
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