Given for first student
`R_(2)=10Omega,R_(1)=5Omega,R_(3)=5Omega`
for second student `R_(1)=500Omega,R_(3)=5Omega`
Now, according to wheatstone bridge rule,
`(R_(2))/(R)=(R_(1))/(R_(3))impliesR=R_(3)xx(R_(2))/(R_(1))`
Now putting all the values inn Eq. (i) we get `R=10Omega` for both student. thus, we can analyse that the wheatstone bridge is most sensitive and accurate if resistance are of same value. thus, the errors of measurement of the two studets depend on te accuracy and sensitivity of the bridge which inturn depends on the accuracy with which `R_(2)` and `R_(1)` can be measured.
When `R_(2)` and `R_(1)` are larger,t he currents though the arms of bridge is very weak. This can make the determination of null point accurately more difficult.