Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion
Discussion Forum : Electronic Devices and Circuits - Section 1 (Q.No. 39)
39.
A potential of 7 V is applied to a silicon diode. A resistance of 1 K ohm is also in series with the diode. The current is
Answer: Option
Explanation:
Discussion:
7 comments Page 1 of 1.
Prasad said:
1 decade ago
How it is possible? Please explain anyone.
Saranya said:
1 decade ago
I = V/R.
V = 7v.
R = 1k.
I = (7-0.7)/1000 = 0.007{The forward voltage drop for Silicon diode is 0.7v}.
i.e I = 6.3/1000 = 6.3mA.
V = 7v.
R = 1k.
I = (7-0.7)/1000 = 0.007{The forward voltage drop for Silicon diode is 0.7v}.
i.e I = 6.3/1000 = 6.3mA.
Thopu said:
8 years ago
@Saranya.
Explain it clearly.
Explain it clearly.
Mononisha.R said:
7 years ago
I=V/R,
where V=V-Vsi,
Vsi=0.7V and V=7V
R=1K ohm = 1000ohm.
So, I=(7-0.7)/1000.
=6.3/1000,
I = 6.3 mA.
where V=V-Vsi,
Vsi=0.7V and V=7V
R=1K ohm = 1000ohm.
So, I=(7-0.7)/1000.
=6.3/1000,
I = 6.3 mA.
Asad said:
5 years ago
Voltage drop should be considered.
Vikas said:
5 years ago
Yes, voltage drop must be considered because here series term is used.
Shanti Bambhaniya said:
4 years ago
I = V/R.
V = 7v.
R = 1k.
I = (7-0.7)/1000 = 0.007{The forward voltage drop for Silicon diode is 0.7v}.
{The forward voltage drop for Germanium diode is 0.3v}
i.e I = 6.3/1000 = 6.3mA.
V = 7v.
R = 1k.
I = (7-0.7)/1000 = 0.007{The forward voltage drop for Silicon diode is 0.7v}.
{The forward voltage drop for Germanium diode is 0.3v}
i.e I = 6.3/1000 = 6.3mA.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers