Sir I have a doubt while calculating the current while the key is plugged in and letting the current pass through shunt resistance, why is the I prime equal to (E/(R+(GS/G+S))) × (S /(G+S)) why did we just multiply that extra quantity of Reciprocals or something
Thank you sir, the video was very useful I was able to understand the procedure but I did not understand theory behind it You made it very clear , I am more confident in my practical now 👍👍👍 Thanks a lot again sir ❤️
Brilliantly explained sir.... In our school we were told to just memorize the formula.... But i always wanted to learn and observe all the formulaes and theory behind them... This also made me clear that why we take only half deflection of the galavanometer when key for shunt resistance is on.... Which is done as to a get us a simple formula for G = RS/R-S.... Otherwise that wouldn't come for sure... But ya i wanted to ask if the figure of merit remains constant for a particular galvanometer.... Bcz if it is constant then only current will get half due to half deflection as I = k(theta) Third and the lastly I wanted to ask sir .... If I/2 current flown in galvanometer after suitable value of shunt resistance ..(assuming before key was on there was I ampere)... Then does that mean exactly I/2 also flow through that shunt resistance.... Because to do that the ratio of G and S must be 1 then only current drawn will be same by both... And if we look in the formula G = RS/R-S.... Then if i approximate by R>>>>>S then G=S.... So can we convey that same I/2 current flows through the shunt resistance and I current through the circuit??? But don't you think doing so will violate the effective resistance ?? Because when key was off then Reff= R+G After key was on then Reff= R + GS/S+G = R+ G/2 This would also change the net current through the circuit...So how can this be possible?? Is this contradiction due to our assumption that we made for G =S? Please answer if possible... It would help me a lot.
First of all, thank you for the interaction For a given galvanometer, ‘the figure of merit’ is constant. In the second part of the experiment, the current through the galvanometer is , no doubt I/2. But the main current in the circuit is now I1. Therefore, the current through the shunt resistance is I1 -(I/2)
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Wonderfully explained sir. Thank you so much. Understood every bit of it clearly.
Thank you for your positive remark
Thanks a lot sir for this helpful video...me and my friend were able to understand the concept really well
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it took me so long to find a video explaining the concept behind the method....
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Thank you sir.!
Great Explanation!!
Nice... Sir
Thank you Pandy
Sir I have a doubt while calculating the current while the key is plugged in and letting the current pass through shunt resistance, why is the I prime equal to (E/(R+(GS/G+S))) × (S /(G+S)) why did we just multiply that extra quantity of Reciprocals or something
Thank u sir for an amazing video,it actually helped me with my practicals.😄😄😄
Thank you so much sir 😌
finally got this topic ... very nice explanation sir
Thank you Anubhav
Thank you sir, the video was very useful
I was able to understand the procedure but I did not understand theory behind it
You made it very clear , I am more confident in my practical now 👍👍👍
Thanks a lot again sir ❤️
Sir you made this so easy for me
Brilliantly explained sir.... In our school we were told to just memorize the formula.... But i always wanted to learn and observe all the formulaes and theory behind them... This also made me clear that why we take only half deflection of the galavanometer when key for shunt resistance is on.... Which is done as to a get us a simple formula for G = RS/R-S.... Otherwise that wouldn't come for sure...
But ya i wanted to ask if the figure of merit remains constant for a particular galvanometer.... Bcz if it is constant then only current will get half due to half deflection as I = k(theta)
Third and the lastly I wanted to ask sir .... If I/2 current flown in galvanometer after suitable value of shunt resistance ..(assuming before key was on there was I ampere)... Then does that mean exactly I/2 also flow through that shunt resistance.... Because to do that the ratio of G and S must be 1 then only current drawn will be same by both... And if we look in the formula G = RS/R-S.... Then if i approximate by R>>>>>S then G=S....
So can we convey that same I/2 current flows through the shunt resistance and I current through the circuit??? But don't you think doing so will violate the effective resistance ??
Because when key was off then
Reff= R+G
After key was on then
Reff= R + GS/S+G = R+ G/2
This would also change the net current through the circuit...So how can this be possible?? Is this contradiction due to our assumption that we made for G =S?
Please answer if possible... It would help me a lot.
First of all, thank you for the interaction
For a given galvanometer, ‘the figure of merit’ is constant.
In the second part of the experiment, the current through the galvanometer is , no doubt I/2. But the main current in the circuit is now I1. Therefore, the current through the shunt resistance is I1 -(I/2)
@@panneerselvam9802 thank you very much sir for your immediate reply☺️
Hey please explain me
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