Question 118: Current Electricity

Question

To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm . The least count of scale is 1 cm . The percentage error in the ratio of EMFs is $\_\_\_\_$

Accepted Answer

The EMF (E) of a cell in a potentiometer is directly proportional to its balancing length (1).
Therefore, the ratio of the EMFs of two cells is given by :
$$ \mathrm{R}=\frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{l}_1}{\mathrm{l}_2} $$
Balancing length for first cell $\mathrm{l}_1=200 \mathrm{~cm}$
Balancing length for second cell $\mathrm{l}_2=150 \mathrm{~cm}$
Absolute error in measurement $\Delta \mathrm{l}_1=\Delta \mathrm{l}_2=1 \mathrm{~cm}$ (the least count of the scale)
For a ratio $\mathrm{R}=\frac{\mathrm{l}_1}{\mathrm{l}_2}$, the relative error $\left(\frac{\Delta \mathrm{R}}{\mathrm{R}}ight)$ is the sum of the relative errors of the individual components :
$$ \frac{\Delta \mathrm{R}}{\mathrm{R}}=\frac{\Delta \mathrm{l}_1}{\mathrm{l}_1}+\frac{\Delta \mathrm{l}_2}{\mathrm{l}_2} $$
Percentage Error $=\left(\frac{\Delta \mathrm{l}_1}{\mathrm{l}_1}+\frac{\Delta \mathrm{l}_2}{\mathrm{l}_2}ight) 	imes 100=\left(\frac{1}{200}+\frac{1}{150}ight) 	imes 100 \approx 1.17 \%$
The calculated percentage error is approximately $1.17 \%$.

Answer

1.75

Answer

1.55

Answer

1.17

Answer

1.65

Current Electricity

To compare EMF of two cells using potentiometer the balancing lengths obtained are 200 cm and 150 cm . The least count of scale is 1 cm . The percentage error in the ratio of EMFs is \_\_\_\_

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