End correction in meter bridge practical

Apr 2017
2
0
Could end correction be minus ? In meter bridge practical , I got minus end correction values .. is it possible ? Can anyone explain me what that means ?
Thanks in advance .
 
Apr 2015
1,035
223
Somerset, England
Hello Madhu and welcome.

It is really good to see some practical work being done, this is just so important.

:)

The answer to your question lies in understanding where the end corrections come from.

I have drawn a greatly exaggerated error to show the issue.

When you read the bridge you measure from the balance point to what you hope are the ends of the slidewire.
The ends of the slidewire are the points where the heavier copper bus bars are attached (eg soldered) on. They may not be exactly 1 metre apart. I have labelled these points M and N. Also the ends of the ruler may not be exactly aligned with these points.

So I have shown two different errors, alpha and beta, one where the ruler sticks out and the other where the wire sticks out further.

Measurements L1 and L2 are shown and the true lengths for the measurements are L1+a and L2+b.
Of course a is positive and b is negative in this example.

Does this help?
 

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Apr 2017
2
0
Thank you.

Thank you very much .. it was really helpful . I couldn't understand the reason for the minus end correction . Because of your explanation , now I understand . (Bigsmile)
 
Apr 2015
1,035
223
Somerset, England
Thank you very much .. it was really helpful . I couldn't understand the reason for the minus end correction . Because of your explanation , now I understand . (Bigsmile)
There is another possible error source, called connection or contact resistance.

Although the bus bars are heavy enough to consider them as having nelgigible (zero) resistance, the solder joint where they conect to the wire may not.

If each joint has a resistance r, and the test resistances are P and Q then measuring from the balance point each arm now sees a resistance of

P + r and Q + r

If we write Q/p = n or Q = nP then we have the ratio of the resistances as

(P + r) / (nP + r) instead of P/Q or P/nP

You might like to think about this as r becomes large compared to P.
 
Nov 2019
1
0
Assam
the negative end correction cannot be possible,

if we measure an unknown small resistance

which is less than 1 ohm then we couldn`t get the perfect resistance of the given wire due to end error of the meter bridge wire connection.
another extra resistance from the left end of the wire will add up.
P = ρl1 +r/
where ρl1 is the resistance of the wire from the left end equal to the resistance connected in the left gab and r/ is the resistance exists at the end due to connection of the wire. This extra resistance adds to the resistance of the wire.
Now if we now that ρl1 is already less than P i.e,

P>ρl1

So if r/ become –ve then decrease ρl1 which gives more or increases the error or the wire resistance.

for example.



Suppose the given resistance p to be measure is = 2 ohm

Now, from meter bridge wire we get ρl1 = 1.5ohm

another resistance is exist at the end connection which is r/

from principle we know that,

P = ρl1 +r/



r/ = p - ρl1
= 2-1.5

r/= 0.5ohm

hence end correction is always positive because resistance is always positive.