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High School/JEE/NEET/IPhO Physics | 17-19 Yrs

Circuit Analysis
Duration: 3.37 Min
 
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Duration: 8.06 Min
 
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Duration: 9.29 Min
 
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Duration: 8.56 Min
 
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Duration: 4.18 Min
 
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Duration: 5.39 Min
 
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Duration: 6.20 Min
 
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Module - 10 Solved Example-4
Duration: 1.52 Min
 
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Module - 11 Solved Example-5
Duration: 2.28 Min
 
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Duration: 5.07 Min
 
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Module - 13 Solved Example-6
Duration: 5.05 Min
 
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Module - 14 Solved Example-7
Duration: 5.05 Min
 
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Module - 15 Solved Example-8
Duration: 4.11 Min
 
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Module - 16 Solved Example-9
Duration: 5.57 Min
 
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Duration: 5.29 Min
 
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Duration: 4.27 Min
 
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Duration: 6.50 Min
 
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Duration: 5.28 Min
 
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Module - 21 Battery Grid
Duration: 2.59 Min
 
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Duration: 7.35 Min
 
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Duration: 2.58 Min
 
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Duration: 3.48 Min
 
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Duration: 4.32 Min
 
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  1.  
  2. in writibg kvl for a-b u did not consider 6ohm resistance why???????????????????????/
    3 years ago by

    Post your answer here

  3.  
  4. sir using kcl im getting 5/14 ampere .by branch mainpulaion i kept 4ohm and 6ohm in series ..finally i got this where im going wrong(did with single varibale)
    5 years ago by Sarju Panchal

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  5.  
  6. Why the 6 ohm resistance is not considered in KVL eqn from A to B? kindly let me know. Regards.
    7 years ago by Ayushi Chaugaonkar
    Ans 1 ->
    because for loop 1 you can see that current I is flowing across 3 ohms in direction opposite to direction we are considering that is from A to B clockwise... so we add 3ohm resistance but if you see for the 6ohm resistance the current I does not flow through that branch and plus A and B is open so we do not consider the potential drop across 6ohm as no current is flowing through it. hope this helps
    7 years ago by ABHISHEK SHANKAR

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  7.  
  8. HOw to solve by KCL method ??? Please sir HELP !!
    7 years ago by Pratyush Ranjan Roul

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  9.  
  10. how can we solve this problem by KCL method??
    8 years ago by

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  11.  
  12. sir i didnt understand why didnt you consider potential drop across the resistance 6
    9 years ago by Aishwarya Roogi
    Ans 1 ->
    right part of the circuit is open so current=0 as per formula ir=0 so potential drop=0
    8 years ago by akanksh

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