Why?

Well, since it is not intuitive to you, I would recommend you tell me what each meter should read. Maybe that will help you solve the problem.

 

Incorrect.

ok

 

Well, since it is not intuitive to you, I would recommend you tell me what each meter should read. Maybe that will help you solve the problem.

ok,
R1 1K 12mA
R2 2K 6mA
R3 3K 4mA
R4 4K 3mA

 

Incredible though it may seem

why were for month outside from USA?

 

Is these calculation are right?

 

Is these calculation are right?

I asked

I would recommend you tell me what each meter should read.

You gave me the current through each resistor.

Review Kirchoff's Laws and apply the applicable law to this circuit. Then answer my question.

p.s. If this circuit were asked on a job interview, I would thank you for your answer and stop the interview, rejecting your application.

 

ok,
R1 1K 12mA
R2 2K 6mA
R3 3K 4mA
R4 4K 3mA

So show HOW you are getting these numbers. They are wrong, but we can't begin to tell you why you are getting the wrong answers unless you show use HOW you are getting them.

Gee, that sounds like an echo.

 

how to know or calculate this.
Which component is shorted?

@RRITESH KAKKAR
That you need even ask such a question tells me you haven't so much as thought about the question! -- Look at the diagram? Apart from the battery, what component do all the indicators have in common?

Hint: The implied 5nΩ should be regarded as characteristic of the battery and/or the indicators -- Come to that - it could all be in the defective resistor given that the indicators are limited to four digit precision -- This is a no brainier -- I've got to wonder if you aren't having a bit of fun at our expense???

Golly! -- These fora are addictive

Best regards
HP

 

Yes, it was far better circuit to understand

yes the circuit is normal working..may be R4 open because there is no current following can be seen.

@RRITESH KAKKAR
Does that make sense? Ask yourself what AM4 would read if R4=∞Ω? --- I assert that you know better... --- How's about you cool the comedy and settle down to serious study? -- Give positive attention a whirl - who knows? You might find that you prefer it

Hopefully
HP

 

Let Calculate the total equivalent Resistance.
1/Req=1/R1+1/R2+1/R3+1/R4
1/Req=1/1kohm+1/2kohm+1/3kohm+1/4kohm
1/Req=1kohm +0.5kohm+0.33kohm+0.25kohm
1/Req=1/2.08kohm
Req=480ohm

so, we know Vt=It*Req
Vt=12V
Req=480ohm
It=25mA

AM1=25mA (AM2+AM3+AM4+IR1)
AM2=25mA-IR1(current through R1)
AM2=25mA-(12/1000)=25mA-12mA=13mA
AM3=13mA-(12/2000)=13mA-6mA=7mA
AM4=7mA-(12/3000)=7mA-4mA=3mA
IR4=2mA

I think there is fault in R4...

 

Review Kirchoff's Laws and apply the applicable law to this circuit. Then answer my question.

Kirchhoff's first law The current entering any junction is equal to the current leaving that junction. i2 + i3 = i1 + i4
Kirchhoff's second law The sum of all the voltages around a loop is equal to zero.
v1 + v2 + v3 - v4 = 0

 

AM1=25mA (AM2+AM3+AM4+IR1)
AM2=25mA-IR1(current through R1)
AM2=25mA-(12/1000)=25mA-12mA=13mA
AM3=13mA-(12/2000)=13mA-6mA=7mA
AM4=7mA-(12/3000)=7mA-4mA=3mA
IR4=2mA

How did you determine R4 had 2 mA through it? I'll assume it was a typo.

There are ONLY 4 ammeters in the circuit. All are reading 12 Amperes. I had asked what they were suppose to read.

The simplest way to complete that task was:

AM4 = IR4
AM3 = AM4 + IR3
AM2 = AM3 + IR2
AM1 = AM2 + IR1

Now you stated

I think there is fault in R4...

Can you expand on the fault? How many ways can a resistor be faulty?

 

Can you expand on the fault? How many ways can a resistor be faulty?

1. May be open
2. May be short
3. May be some time resistance increase or decrease due to ages.

 

Ok ...which fault condition caused the readings you saw on the diagram?

 

Let Calculate the total equivalent Resistance.
1/Req=1/R1+1/R2+1/R3+1/R4
1/Req=1/1kohm+1/2kohm+1/3kohm+1/4kohm
1/Req=1kohm +0.5kohm+0.33kohm+0.25kohm
1/Req=1/2.08kohm
Req=480ohm

so, we know Vt=It*Req
Vt=12V
Req=480ohm
It=25mA

AM1=25mA (AM2+AM3+AM4+IR1)
AM2=25mA-IR1(current through R1)
AM2=25mA-(12/1000)=25mA-12mA=13mA
AM3=13mA-(12/2000)=13mA-6mA=7mA
AM4=7mA-(12/3000)=7mA-4mA=3mA
IR4=2mA

I think there is fault in R4...

What would each of the meters read if R1 was, say, 1 Ω and all of the others were, say, 1 kΩ?
What would each of the meters read if R2 was, say, 1 Ω and all of the others were, say, 1 kΩ?
What would each of the meters read if R3 was, say, 1 Ω and all of the others were, say, 1 kΩ?
What would each of the meters read if R4 was, say, 1 Ω and all of the others were, say, 1 kΩ?

Which, if any, of these results are somewhat similar to what the meters are showing in the problem?

 

@RRITESH KAKKAR
You have previously demonstrated ability to solve this sort of 'problem' on several occasions (CIP phase shift in X+R and X||R circuits)! - Moreover you have shown proficiency with LCR impedance transformation networks and Z-form conversion!

Could you please post links to threads where he demonstrates these capabilities?

 

Could you please post links to threads where he demonstrates these capabilities?

He demonstrated skill at manipulating vectors (Ca post 700+ on his thread in the math forum) once he understood the question -- Moreover, he responded correctly to pool exercises in regards to phase shift attendant to impedances comprised of various reactance-resistance ratios/orientations (IIRC on a thread in the 'general chat' forum) the fora's 'search' feature may help...

FWIW: The intent of my challenges to his sincerity is less that of accusation than an attempt to 'rudely' acquaint him with the fact that he is in possession of the requisite skills - and, hence, quite capable of performing the exercises... While I sympathize with the desire to remain within one's personal 'comfort zone' -- I likewise recognize the fact that education (indeed, 'growth' in general) and 'comfort' are, not uncommonly, in diametric opposition... Point being: My intent is that of encouragement not harassment! --- Once again, I am not and never have been professionally employed as in instructor - neither do I claim proficiency or expertise at said 'calling' -- That said - it has been my observation that the distinction between "special" and "gifted" is frequently down to discipline (CIP concentration) and 'daring' alone -- whether 'innate' or 'acquired' -- The result is the same!!!

Many thanks for your concern
HP

 

What would each of the meters read if R1 was, say, 1 Ω and all of the others were, say, 1 kΩ?

AM4=0.012A
AM3=0.012A+0.012A=0.024A
AM2=0.024A+0.012A=0.036A
AM1=0.036A+12A=12.036A

 

What would each of the meters read if R2 was, say, 1 Ω and all of the others were, say, 1 kΩ?

AM4=0.012A
AM3=0.012A+0.012A=0.024A
AM2=0.024A+12A=12.024A
AM1=12.024A+0.012A=12.036A

 

What would each of the meters read if R3 was, say, 1 Ω and all of the others were, say, 1 kΩ?

AM4=0.012A
AM3=0.012A+12A=12.012A
AM2=12.012A+0.012A=12.024A
AM1=12.024A+0.012A=12.036A