Help With Resistor Type? [on hold]
up vote
10
down vote
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Im having a hard time finding the spects for this piece
I know is a metal film resistor, according to the colors its a 470? is this correct, does anybody know where can I purchase a ton of this?
resistors
asked yesterday
G.Medina
574
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G.Medina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put on hold as off-topic by Dave Tweed♦ 15 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Questions seeking recommendations for specific products or places to purchase them are off-topic as they are rarely useful to others and quickly obsolete. Instead, describe your situation and the specific problem you're trying to solve." – Dave Tweed
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
10
down vote
favorite
Im having a hard time finding the spects for this piece
I know is a metal film resistor, according to the colors its a 470? is this correct, does anybody know where can I purchase a ton of this?
resistors
asked yesterday
G.Medina
574
New contributor
G.Medina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Dave Tweed♦ 15 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Questions seeking recommendations for specific products or places to purchase them are off-topic as they are rarely useful to others and quickly obsolete. Instead, describe your situation and the specific problem you're trying to solve." – Dave Tweed
If this question can be reworded to fit the rules in the help center, please edit the question.
8
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
yesterday
7
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
yesterday
add a comment |
up vote
10
down vote
favorite
up vote
10
down vote
favorite
Im having a hard time finding the spects for this piece
I know is a metal film resistor, according to the colors its a 470? is this correct, does anybody know where can I purchase a ton of this?
resistors
asked yesterday
G.Medina
574
New contributor
G.Medina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Im having a hard time finding the spects for this piece
I know is a metal film resistor, according to the colors its a 470? is this correct, does anybody know where can I purchase a ton of this?
resistors
resistors
asked yesterday
G.Medina
574
New contributor
G.Medina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
G.Medina
574
New contributor
G.Medina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
G.Medina
574
New contributor
G.Medina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
G.Medina
574
asked yesterday
G.Medina
574
574
New contributor
G.Medina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
G.Medina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
G.Medina is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by Dave Tweed♦ 15 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Questions seeking recommendations for specific products or places to purchase them are off-topic as they are rarely useful to others and quickly obsolete. Instead, describe your situation and the specific problem you're trying to solve." – Dave Tweed
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Dave Tweed♦ 15 hours ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Questions seeking recommendations for specific products or places to purchase them are off-topic as they are rarely useful to others and quickly obsolete. Instead, describe your situation and the specific problem you're trying to solve." – Dave Tweed
If this question can be reworded to fit the rules in the help center, please edit the question.
8
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
yesterday
7
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
yesterday
add a comment |
8
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
yesterday
7
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
yesterday
8
8
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
yesterday
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
yesterday
7
7
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
yesterday
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
33
down vote
So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.
The hard part really is
where can I purchase a ton of this
because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.
That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.
So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.
I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.
A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is
$$E=frac12 LI^2text,$$
leading to a total maximum energy stored in your ton of inductors of
begin{align}
E_{tot}&=Ncdotfrac12 LI^2\
&=leftlceilfrac{1,text{Mg}}{185,text{mg}} rightrceilcdot frac12 47,mathrmmutext{H},left(205,text{mA}right)^2\
&approx 5.4cdot 10^6,cdot, 2.4cdot10^{-6} ,text{kg},text{m}^2,text{s}^{−2},text{A}^{−2},cdot ,4.2cdot10^{-3},text{A}^2\
&=5.4cdot2.4cdot4.2cdot10^{-3},text{N·m}\
&approx 54 ,text{J.}
end{align}
That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.
answered yesterday
Marcus Müller
30k35691
13
Again with the dry comedy!
– winny
yesterday
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
yesterday
2
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
yesterday
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
yesterday
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
yesterday
|
show 1 more comment
up vote
11
down vote
NO, it is an inductor -- you can see on PCB they define it as L1
the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%
answered yesterday
Satish Singupuram
3958
New contributor
Satish Singupuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I think you mean 47uH, right?
– Spehro Pefhany
yesterday
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
yesterday
the 3rd band is brown.. i think.. if it is brown then it is 470uH, if it is black then 47uH
– Satish Singupuram
yesterday
1
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
yesterday
1
@SpehroPefhany, its brown actually
– Satish Singupuram
yesterday
|
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
33
down vote
So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.
The hard part really is
where can I purchase a ton of this
because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.
That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.
So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.
I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.
A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is
$$E=frac12 LI^2text,$$
leading to a total maximum energy stored in your ton of inductors of
begin{align}
E_{tot}&=Ncdotfrac12 LI^2\
&=leftlceilfrac{1,text{Mg}}{185,text{mg}} rightrceilcdot frac12 47,mathrmmutext{H},left(205,text{mA}right)^2\
&approx 5.4cdot 10^6,cdot, 2.4cdot10^{-6} ,text{kg},text{m}^2,text{s}^{−2},text{A}^{−2},cdot ,4.2cdot10^{-3},text{A}^2\
&=5.4cdot2.4cdot4.2cdot10^{-3},text{N·m}\
&approx 54 ,text{J.}
end{align}
That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.
answered yesterday
Marcus Müller
30k35691
13
Again with the dry comedy!
– winny
yesterday
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
yesterday
2
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
yesterday
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
yesterday
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
yesterday
|
show 1 more comment
up vote
33
down vote
So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.
The hard part really is
where can I purchase a ton of this
because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.
That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.
So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.
I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.
A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is
$$E=frac12 LI^2text,$$
leading to a total maximum energy stored in your ton of inductors of
begin{align}
E_{tot}&=Ncdotfrac12 LI^2\
&=leftlceilfrac{1,text{Mg}}{185,text{mg}} rightrceilcdot frac12 47,mathrmmutext{H},left(205,text{mA}right)^2\
&approx 5.4cdot 10^6,cdot, 2.4cdot10^{-6} ,text{kg},text{m}^2,text{s}^{−2},text{A}^{−2},cdot ,4.2cdot10^{-3},text{A}^2\
&=5.4cdot2.4cdot4.2cdot10^{-3},text{N·m}\
&approx 54 ,text{J.}
end{align}
That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.
answered yesterday
Marcus Müller
30k35691
13
Again with the dry comedy!
– winny
yesterday
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
yesterday
2
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
yesterday
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
yesterday
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
yesterday
|
show 1 more comment
up vote
33
down vote
up vote
33
down vote
So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.
The hard part really is
where can I purchase a ton of this
because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.
That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.
So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.
I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.
A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is
$$E=frac12 LI^2text,$$
leading to a total maximum energy stored in your ton of inductors of
begin{align}
E_{tot}&=Ncdotfrac12 LI^2\
&=leftlceilfrac{1,text{Mg}}{185,text{mg}} rightrceilcdot frac12 47,mathrmmutext{H},left(205,text{mA}right)^2\
&approx 5.4cdot 10^6,cdot, 2.4cdot10^{-6} ,text{kg},text{m}^2,text{s}^{−2},text{A}^{−2},cdot ,4.2cdot10^{-3},text{A}^2\
&=5.4cdot2.4cdot4.2cdot10^{-3},text{N·m}\
&approx 54 ,text{J.}
end{align}
That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.
answered yesterday
Marcus Müller
30k35691
So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.
The hard part really is
where can I purchase a ton of this
because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.
That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.
So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.
I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.
A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is
$$E=frac12 LI^2text,$$
leading to a total maximum energy stored in your ton of inductors of
begin{align}
E_{tot}&=Ncdotfrac12 LI^2\
&=leftlceilfrac{1,text{Mg}}{185,text{mg}} rightrceilcdot frac12 47,mathrmmutext{H},left(205,text{mA}right)^2\
&approx 5.4cdot 10^6,cdot, 2.4cdot10^{-6} ,text{kg},text{m}^2,text{s}^{−2},text{A}^{−2},cdot ,4.2cdot10^{-3},text{A}^2\
&=5.4cdot2.4cdot4.2cdot10^{-3},text{N·m}\
&approx 54 ,text{J.}
end{align}
That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.
answered yesterday
Marcus Müller
30k35691
edited yesterday
answered yesterday
Marcus Müller
30k35691
answered yesterday
Marcus Müller
30k35691
answered yesterday
Marcus Müller
30k35691
30k35691
13
Again with the dry comedy!
– winny
yesterday
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
yesterday
2
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
yesterday
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
yesterday
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
yesterday
|
show 1 more comment
13
Again with the dry comedy!
– winny
yesterday
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
yesterday
2
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
yesterday
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
yesterday
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
yesterday
13
13
Again with the dry comedy!
– winny
yesterday
Again with the dry comedy!
– winny
yesterday
2
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
yesterday
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
yesterday
2
2
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
yesterday
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
yesterday
2
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
yesterday
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
yesterday
1
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
yesterday
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
yesterday
|
show 1 more comment
up vote
11
down vote
NO, it is an inductor -- you can see on PCB they define it as L1
the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%
answered yesterday
Satish Singupuram
3958
New contributor
Satish Singupuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I think you mean 47uH, right?
– Spehro Pefhany
yesterday
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
yesterday
the 3rd band is brown.. i think.. if it is brown then it is 470uH, if it is black then 47uH
– Satish Singupuram
yesterday
1
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
yesterday
1
@SpehroPefhany, its brown actually
– Satish Singupuram
yesterday
|
show 2 more comments
up vote
11
down vote
NO, it is an inductor -- you can see on PCB they define it as L1
the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%
answered yesterday
Satish Singupuram
3958
New contributor
Satish Singupuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I think you mean 47uH, right?
– Spehro Pefhany
yesterday
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
yesterday
the 3rd band is brown.. i think.. if it is brown then it is 470uH, if it is black then 47uH
– Satish Singupuram
yesterday
1
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
yesterday
1
@SpehroPefhany, its brown actually
– Satish Singupuram
yesterday
|
show 2 more comments
up vote
11
down vote
up vote
11
down vote
NO, it is an inductor -- you can see on PCB they define it as L1
the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%
answered yesterday
Satish Singupuram
3958
New contributor
Satish Singupuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
NO, it is an inductor -- you can see on PCB they define it as L1
the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%
answered yesterday
Satish Singupuram
3958
New contributor
Satish Singupuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Satish Singupuram
3958
New contributor
Satish Singupuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered yesterday
Satish Singupuram
3958
answered yesterday
Satish Singupuram
3958
3958
New contributor
Satish Singupuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Satish Singupuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Satish Singupuram is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
I think you mean 47uH, right?
– Spehro Pefhany
yesterday
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
yesterday
the 3rd band is brown.. i think.. if it is brown then it is 470uH, if it is black then 47uH
– Satish Singupuram
yesterday
1
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
yesterday
1
@SpehroPefhany, its brown actually
– Satish Singupuram
yesterday
|
show 2 more comments
1
I think you mean 47uH, right?
– Spehro Pefhany
yesterday
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
yesterday
the 3rd band is brown.. i think.. if it is brown then it is 470uH, if it is black then 47uH
– Satish Singupuram
yesterday
1
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
yesterday
1
@SpehroPefhany, its brown actually
– Satish Singupuram
yesterday
1
1
I think you mean 47uH, right?
– Spehro Pefhany
yesterday
I think you mean 47uH, right?
– Spehro Pefhany
yesterday
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
yesterday
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
yesterday
the 3rd band is brown.. i think.. if it is brown then it is 470uH, if it is black then 47uH
– Satish Singupuram
yesterday
the 3rd band is brown.. i think.. if it is brown then it is 470uH, if it is black then 47uH
– Satish Singupuram
yesterday
1
1
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
yesterday
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
yesterday
1
1
@SpehroPefhany, its brown actually
– Satish Singupuram
yesterday
@SpehroPefhany, its brown actually
– Satish Singupuram
yesterday
|
show 2 more comments
8
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
yesterday
7
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
yesterday