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The measurement of an unknown resistance $\mathrm{R}$ is to be carried out using. Wheatstones bridge as given in the figure below. Two students perform an experiment in two ways. Thefirst studentstakes $R_2=10 \Omega$ and $R_1=5 \Omega$. The other student takes $\mathrm{R}_2=1000 \Omega$ and $\mathrm{R}_1=500 \Omega$. In the standard arm, both take $\mathrm{R}_3=5 \Omega$.
Both find $\mathrm{R}=\frac{\mathrm{R}_2}{\mathrm{R}_1}, \mathrm{R}_3=10 \Omega$ within errors.
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Both find $\mathrm{R}=\frac{\mathrm{R}_2}{\mathrm{R}_1}, \mathrm{R}_3=10 \Omega$ within errors.
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The correct answers are:
Errors of measurement do depend on the accuracy with which $\mathrm{R}_2$ and $\mathrm{R}_1$ can be measured
,
If the student uses large values of $R_2$ and $R_1$ the currents through the arms will be feeble. This will make determination of null point accurately more difficult
Errors of measurement do depend on the accuracy with which $\mathrm{R}_2$ and $\mathrm{R}_1$ can be measured
,
If the student uses large values of $R_2$ and $R_1$ the currents through the arms will be feeble. This will make determination of null point accurately more difficult
As given that,
Resistance for first student, $\mathrm{R}_1=5 \Omega, \mathrm{R}_2=10 \Omega, \mathrm{R}_3=5 \Omega$
Resistance for second student, $\mathrm{R}_1=500 \Omega$, $\mathrm{R}_2=1000 \Omega, \mathrm{R}_3=5 \Omega$
Now, according to Wheatstone bridge balanced condition, $\frac{\mathrm{R}_2}{\mathrm{R}}=\frac{\mathrm{R}_1}{\mathrm{R}_3} \quad$ or $\quad \mathrm{R}=\mathrm{R}_3 \times \frac{\mathrm{R}_2}{\mathrm{R}_1}$
Now putting all the values in Eq. (i), we get $R=10 \Omega$ for both students. Thus, we can analyse that the Wheatstone bridge is most sensitive and accurate if resistances are of same value. So the results of both students depend on the accuracy of resistances used.
When $R_2$ and $R_1$ are larger, the currents through galvanometer becomes very weak. It can make the determination of null point accurately more complex.
Resistance for first student, $\mathrm{R}_1=5 \Omega, \mathrm{R}_2=10 \Omega, \mathrm{R}_3=5 \Omega$
Resistance for second student, $\mathrm{R}_1=500 \Omega$, $\mathrm{R}_2=1000 \Omega, \mathrm{R}_3=5 \Omega$
Now, according to Wheatstone bridge balanced condition, $\frac{\mathrm{R}_2}{\mathrm{R}}=\frac{\mathrm{R}_1}{\mathrm{R}_3} \quad$ or $\quad \mathrm{R}=\mathrm{R}_3 \times \frac{\mathrm{R}_2}{\mathrm{R}_1}$
Now putting all the values in Eq. (i), we get $R=10 \Omega$ for both students. Thus, we can analyse that the Wheatstone bridge is most sensitive and accurate if resistances are of same value. So the results of both students depend on the accuracy of resistances used.
When $R_2$ and $R_1$ are larger, the currents through galvanometer becomes very weak. It can make the determination of null point accurately more complex.
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