Question

The length of a potentiometer wire is 600 cm and it carries a current of 40 mA. for a cell of emf 2V and internal resistance 10ohm, the null point is found to be at 500 cm. if a voltmeter is connected across the cell, the balancing length is decreased by 10 cm. find (i) the resistance of whole wire (ii) reading of voltmeter, and (iii) resistance of voltmeter

Answer 1

The solution is attached below

0e38885d2bf99f32621c364614243228.pdf

Answer 2

Answer:

(i) The resistance of whole wire is 60 ohms

(ii) The reading of voltmeter is 1.96 V

(iii) The resistance of voltmeter is 490 ohms.

Solution:

The given values are the  

Length of potentiometer $\mathrm{L}_{1}=600 \mathrm{cm}$

Current I = 40 mA

Emf, E = 2 V

Internal Resistance r = 10 ohm

Null point $\mathrm{L}_{\mathrm{n}}=500 \mathrm{cm}$

The null point is the position at which the potentiometer will be balanced.

In a potentiometer we know that

$\frac{R}{r^{\prime}}=\frac{L_{1}}{L_{2}} \rightarrow(1)$

Here R is the complete resistance of wire and r’ is the resistance of the wire at null point, $\mathrm{L}_{1} \text { and } \mathrm{L}_{2}$ are the whole length of wire and null point of the wire respectively.

The resistance at null point of the whole wire can be determined using the Ohms Law,  

$r^{\prime}=\frac{E}{I}=\frac{2}{40 \times 10^{-3}}=\frac{2 \times 1000}{40}=50 \Omega$

Then the resistance of the complete wire from equation (1) is  

$R=r^{\prime} \times \frac{L_{1}}{L_{2}}=50 \times \frac{600}{500}=60 \Omega$

Thus the complete resistance is 60 ohms.

The modified voltage when current is applied in the potentiometer will give the reading of the voltmeter. We know the relation of potentials in potentiometer as  

$\frac{V}{E}=\frac{L_{3}}{L_{2}} \rightarrow(2)$

Here V is the voltmeter reading, E is the emf, $\mathrm{L}_{3}$ is the modified null point or balancing length after current is passed and $\mathrm{L}_{2}$ is the original null point or balancing length.

It is given in the question that balancing length will be decreased by 10 cm when circuit is connected. So, $\mathrm{L}_{3}=500-10=490 \mathrm{cm}$.

Thus from equation (2), we get

$\frac{V}{2}=\frac{490}{500}$

$V=2 \times \frac{490}{500}=1.96 \mathrm{V}$

Thus, the reading in the voltmeter when circuit is connected is 1.96 V.

Let $\mathrm{Rv}$ be the resistance of voltmeter. The relation between the internal resistance and resistance of voltmeter is  

$r=\left(\frac{E}{V}-1\right) R_{V}$

Substitute equation (2) in the above equation, we get

$r=\left(\frac{L_{2}}{L_{3}}-1\right) R_{V}$

$10=\left(\frac{500}{490}-1\right) R_{V}$

$10 \times 490=(500-490) R_{V}$

$\frac{4900}{10}=R_{V}$

Thus,

$\bold{R_{V}=490\ \Omega}$