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.
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.
Manoj said:
1 decade ago
Actually the diode is in ideal condition so there is no cut in voltage (0v). Answer should be 1mA.
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.
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.
Sandeep kumar gupta 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
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
Geetanjali said:
7 years ago
According to ohms law,
I = V/R.
So,
I=12/12000.
= 0.001A.
that is, 1ma.
I = V/R.
So,
I=12/12000.
= 0.001A.
that is, 1ma.
(1)
Devi said:
1 decade ago
Usually we take LED drop voltage as 1.5 . Why here as 0.7 ?
Jan said:
1 decade ago
Answer is A, b/c voltage drop across an ideal diode is 0v.
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