Potentiometer

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1. 1.↑ back to the question

   i. The emfs of two cells can be compared using sum and difference method. ii. When two cells are connected such that the positive terminal of the first cell is connected to the negative terminal of the second cell, the emf of the two cells are added up and the effective emf of the combination of two cells is $E_1+E_2$. This method of connecting two cells is called the sum method. iii. When two cells are connected such that their negative terminals are together or their positive terminals are connected together, then their emf oppose each other and effective emf of the combination of two cells is $E_1-E_2$ (Considering $E_1>E_2$). This method of connecting two cells is called the difference method. iv. Circuit for sum and difference method is connected as shown in below figure. When keys $K_1$ and $K_3$ are closed the cells $E_1$ and $E_2$ are in the sum mode. The null point is obtained using the jockey. v. Let $l_1$ be the length of the wire between the null point and the point A. This corresponds to the emf $(E_1+E_2)$. $$\therefore E_1+E_2=Kl_1\qquad....(1)$$ where K is the potential gradient along the potentiometer wire. vi. Now the key $K_1$ and $K_3$ are kept open and keys $K_2$ and $K_4$ are closed. In this case the two cells are in the difference mode. vii. Again the null point is obtained. Let $l_2$ be the length of the wire between the null point and the point A. This corresponds to emf $(E_1-E_2)$. $$\therefore E_1-E_2=Kl_2\qquad....(2)$$ viii. Dividing equation (1) by equation (2), $$\dfrac{E_1+E_2}{E_1-E_2}=\dfrac{l_1}{l_2}$$ By componendo and dividendo method, $$\dfrac{(E_1+E_2)+(E_1-E_2)}{(E_1+E_2)-(E_1-E_2)}=\dfrac{l_1+l_2}{l_1-l_2}$$ $$\therefore \dfrac{E_1}{E_2}=\dfrac{l_1+l_2}{l_1-l_2}$$ Thus, emf of two cells can be compared using sum and difference method.

2. 2.↑ back to the question

   i. A potentiometer circuit is set up by connecting a battery of emf E, with a key K and a rheostat such that point A is at higher potential than point B. ii. The cells whose emfs are to be compared are connected with their positive terminals at point A and negative terminals to the extreme terminals of a two way key $K_1$ and $K_2$. iii. The central terminal of the two ways key is connected to a galvanometer. The other end of the galvanometer is connected to a jockey (J). iv. Key K is closed and then, key $K_1$ is closed and key $K_2$ is kept open. Therefore, the cell of emf $E_1$ comes into circuit. v. The null point is obtained by touching the jockey at various points on the potentiometer wire AB. vi. Let $l_1$ be the length of the wire between the null point and the point A. Here, $l_1$ corresponds to emf $E_1$ of the cell. Therefore, $$E_1=Kl_1\qquad....(1)$$ where K is the potential gradient along the potentiometer wire. vii. Now key $K_1$ is kept open and key $K_2$ is closed. The cell of emf $E_2$ now comes in the circuit. Again, the null point is obtained with the help of the Jockey. viii. Let $l_2$ be the length of the wire between the null point and point A. Here $l_2$ corresponds to the emf $E_2$ of the cell. $$\therefore E_2=Kl_2\qquad....(2)$$ ix. Dividing equation (1) by equation (2), $$\dfrac{E_1}{E_2}=\dfrac{l_1}{l_2}$$ Thus, emfs of the two cells can be compared and if any one of the emfs is known, the other can be determined.

3. 3.↑ back to the question

   Principle: The potential difference between two points on the wire is directly proportional to the length of the wire between them provided the wire is of uniform cross section, the current through the wire is the same and temperature of the wire remains constant. Explanation: i. A potentiometer consists of a long wire AB of length L and resistance R having uniform cross sectional area A. ii. A cell of emf E having internal resistance r is connected across AB as shown in the figure. iii. When the circuit is switched on, current I pass through the wire. Current through AB, $I=\dfrac{E}{R+r}\qquad....(1)$ iv. Potential difference across AB, $V_{AB}=IR$ $$V_{AB}=\dfrac{ER}{(R+r)}\qquad....[\text{From equation (1)}]$$ v. Therefore, the potential difference per unit length of the wire is, $$\dfrac{V_{AB}}{L}=\dfrac{ER}{L(R+r)}$$ As long as E remains constant, $\dfrac{V_{AB}}{L}$ will remain constant. vi. $\dfrac{V_{AB}}{L}$ is known as potential gradient along AB and is denoted by K. Potential gradient can be defined as potential difference per unit length of wire. vii. Consider a point C on the wire at distance $l$ from the point A, as shown in the figure. If the potential difference between A and C is $V_{AC}$, then, $V_{AC}=Kl$ i.e., $V_{AC}\propto l$ viii. Thus, the potential difference between two points on the wire is directly proportional to the length of the wire between the two points, provided (a) the wire is of uniform cross section, (b) the current through the wire is the same and (c) temperature of the wire remains constant.

4. 4.↑ back to the question

   i. Potentiometer is more sensitive than a voltmeter. ii. A potentiometer can be used to measure a potential difference as well as an emf of a cell. A voltmeter always measures terminal potential difference, and as it draws some current, it cannot be used to measure an emf of a cell. iii. Measurement of potential difference or emf is very accurate in the case of a potentiometer. A very small potential difference of the order $10^{-6}$ volt can be measured with it. Least count of a potentiometer is much better compared to that of a voltmeter. Due to all these reasons potentiometer is preferred over a voltmeter for measuring emf.

5. 5.↑ back to the question

   No.	Potentiometer	Voltmeter
   i.	Its resistance is infinite.	Its resistance is high but finite.
   ii.	It does not draw any current from the source of known e.m.f.	It draws some current from the source of e.m.f.
   iii.	The potential difference measured by it is equal to actual potential difference (p.d.).	The potential difference measured by it is less than the actual potential difference (p.d.).
   iv.	It has high sensitivity.	It has low sensitivity.
   v.	It measures e.m.f as well as p.d.	It measures only p.d.
   vi.	It is used to measure internal resistance of a cell.	It cannot be used to measure the internal resistance of a cell.
   vii.	It is more accurate.	It is less accurate.
   viii.	It does not give direct reading.	It gives direct reading.
   ix.	It is not portable.	It is portable.
   x.	It is used to measure lower voltage values only.	It is used to measure lower as well as higher voltage values.

   [Any four differences]

6. 6.↑ back to the question

   Formula: $\dfrac{V_{AB}}{L}=\dfrac{ER}{L(R+r)}$ Where, $\dfrac{V_{AB}}{L}$ is the potential gradient and E is the electric field.

7. 7.↑ back to the question

   i. The experimental set up for this method consists of a potentiometer wire AB connected in series with a cell of emf E, the key $K_1$, and rheostat as shown in figure. ii. The terminal A is at higher potential than terminal B. A cell of emf $E_1$ whose internal resistance r is to be determined is connected to the potentiometer wire through a galvanometer G and the jockey J. iii. A resistance box R is connected across the cell $E_1$ through the key $K_2$. The key $K_1$ is closed and $K_2$ is open. iv. The circuit now consists of the cell E, cell $E_1$, and the potentiometer wire. The null point is then obtained. v. Let $l_1$ be length of the potentiometer wire between the null point and the point A. This length corresponds to emf $E_1$. $$\therefore E_1=Kl_1\qquad....(1)$$ where K is potential gradient of the potentiometer wire which is constant. vi. Now both the keys $K_1$ and $K_2$ are closed so that the circuit consists of the cell E, the cell $E_1$, the resistance box, the galvanometer and the jockey. Some resistance R is selected from the resistance box and null point is obtained. vii. The length of the wire $l_2$ between the null point and point A is measured. This corresponds to the voltage between the null point and point A. $$\therefore V=Kl_2\qquad....(2)$$ Dividing equation (1) by equation (2), $$\dfrac{E_1}{V}=\dfrac{Kl_1}{Kl_2}=\dfrac{l_1}{l_2}\qquad....(3)$$ viii. Consider the loop PQSTP, $E_1=IR+Ir$ and $V=IR$ $$\therefore \dfrac{E_1}{V}=\dfrac{IR+Ir}{IR}=\dfrac{R+r}{R}=\dfrac{l_1}{l_2}\qquad....[\text{From equation (3)}]$$ $$\Rightarrow r=R\left(\dfrac{l_1}{l_2}-1\right)$$ The above equation is used to determine the internal resistance of the cell.

8. 8.↑ back to the question

   Potential gradient is defined as potential difference per unit length of wire.

9. 9.↑ back to the question

   Given: L = 10 m, R = 10 Ω, I = 0.4 A, $E_1=1.3$ V, $E_2=1.1$ V To find: i. Null point distance when two cells assist each other ($l_1$) ii. Null point distance when two cells oppose each other ($l_2$) Formulae: i. $E_1+E_2=Kl_1$ ii. $E_1-E_2=Kl_2$ Calculation: i. Since, $K=\dfrac{ER}{(R+r)L}=\dfrac{(IR)R}{(R+r)L}$ Here, r = 0 Ω $$\therefore K=\dfrac{IR}{L}=\dfrac{0.4\times10}{10}=0.4\text{ V/m}$$ When $E_1$ and $E_2$ are connected so as to assist then, From formula (i), $$l_1=\dfrac{E_1+E_2}{K}=\dfrac{1.3+1.1}{0.4}$$ $\therefore l_1=6$ m ii. From formula (ii), $l_2=\dfrac{E_1-E_2}{K}=\dfrac{1.3-1.1}{0.4}$ $\therefore l_2=0.5$ m Ans: i. The null point distance when two cells assist each other is 6 m. ii. The null point distance when two cells oppose each other is 0.5 m.

10. 10.↑ back to the question

    Given: $R_1=8\ \Omega$, $l_2=200$ cm, $R_2=4\ \Omega$, $l'_2=160$ cm To find: Balancing length ($l_1$), Internal resistance of the cell (r) Formula: $r=R\left(\dfrac{l_1-l_2}{l_2}\right)$ Calculation: From first condition, $r=8\left(\dfrac{l_1-200}{200}\right)\qquad....(i)$ From second condition, $r=4\left(\dfrac{l_1-160}{160}\right)\qquad....(ii)$ From equation (i) and (ii), $$8\left(\dfrac{l_1-200}{200}\right)=4\left(\dfrac{l_1-160}{160}\right)$$ $$\dfrac{l_1-200}{25}=\dfrac{l_1-160}{40}$$ $25l_1-4000=40l_1-8000$ $15l_1=4000$ $l_1=266.7$ cm Using equation (i) Internal resistance, $$r=8\left(\dfrac{l_1-200}{200}\right)=\dfrac{266.7-200}{25}=\dfrac{66.7}{25}$$ $$=\text{antilog}\{\log(66.7)-\log(25)\}=\text{antilog}\{1.8241-1.3979\}=\text{antilog}\{0.4262\}$$ $\therefore r=2.668\ \Omega$ Ans: i. The balancing length is 266.7 cm. ii. The internal resistance of the cell is 2.668 Ω.

11. 11.↑ back to the question

    Given: L = 4 m, $R_1=5\ \Omega$, E = 2 V, r = 1 Ω, K = $10^{-3}$ V/cm = $10^{-1}$ V/m To find: Series resistance (X) Formula: $K=\dfrac{V}{L}=\dfrac{IR}{L}=\left(\dfrac{E}{R+r+X}\right)\dfrac{R}{L}$ Calculation: From formula, $$10^{-1}=\left(\dfrac{2}{5+1+X}\right)\left(\dfrac{5}{4}\right)$$ $\therefore 6+X=20\times\dfrac{5}{4}$ $\therefore 6+X=25$ $\therefore X=25-6=19\ \Omega$ Ans: A resistance of 19 Ω should be connected in series.

12. 12.↑ back to the question

    Given: σ = 0.1 Ω/cm = 0.1×100 Ω/m = 10 Ω/m, $l_1=300$ cm = 3 m, $E_1=1.5$ V, $E_2=1.4$ V To find: Current (I), balancing length ($l_2$) Formulae: i. $I=\dfrac{E_1}{\sigma l_1}$ ii. $\dfrac{E_1}{E_2}=\dfrac{l_1}{l_2}$ Calculation: Using formula (i), $$I=\dfrac{1.5}{10\times3}=0.05\text{ A}$$ Using formula (ii), $$l_2=\dfrac{E_2l_1}{E_1}=\dfrac{1.4\times3}{1.5}=2.8\text{ m}$$ Ans: i. The flow of current is 0.05 A. ii. The balancing length for second cell is 2.8 m.

13. 13.↑ back to the question

    Given: σ = 0.1 Ω/m, E = 1.5 V, L = 300 cm = 3 m To find: Current through potentiometer (I) Formulae: i. $\sigma=R/L$ ii. $I=E/R$ Calculation: From formula (i), $R=\sigma L=0.1\times3=0.3\ \Omega$ From formula (ii), $I=\dfrac{1.5}{0.3}=5$ A Ans: Current through the potentiometer wire is 5 A.

14. 14.↑ back to the question

    Given: L = 2 m, R = 10 Ω, $R_E=990\ \Omega$, E = 2 V To find: Potential gradient of wire (K) Formula: $K=\dfrac{V}{L}$ Calculation: Since, $I=\dfrac{E}{R+R_E}$ Also, $V=IR=\dfrac{ER}{R+R_E}=\dfrac{2\times10}{10+990}=\dfrac{20}{1000}$ V = $2\times10^{-2}$ volt From formula, $K=\dfrac{2\times10^{-2}}{2}=10^{-2}$ V/m Ans: The potential gradient of wire is $10^{-2}$ V/m.

15. 15.↑ back to the question

    Given: $R_1=8\ \Omega$, $l_2=160$ cm, $R_2=10\ \Omega$, $l'_2=180$ cm To find: Balancing length ($l_1$), Internal resistance (r) Formula: $r=R\left(\dfrac{l_1-l_2}{l_2}\right)$ Calculation: From first condition, $r=8\left(\dfrac{l_1-160}{160}\right)\qquad....(i)$ From second condition, $r=10\left(\dfrac{l_1-180}{180}\right)\qquad....(ii)$ From equation (i) and (ii), $$8\left(\dfrac{l_1-160}{160}\right)=10\left(\dfrac{l_1-180}{180}\right)$$ $$\dfrac{l_1-160}{20}=\dfrac{l_1-180}{18}$$ $\therefore 20l_1-3600=18l_1-2880$ $\therefore 2l_1=720$ $\therefore l_1=360$ cm Using equation (i) we get Internal resistance, $$r=8\left(\dfrac{l_1-160}{160}\right)=\dfrac{360-160}{20}=\dfrac{200}{20}$$ $\therefore r=10\ \Omega$ Ans: i. The balancing length is 360 cm. ii. The internal resistance of the cell is 10 Ω.

16. 16.↑ back to the question

    Given: $E_1=1.5$ V, $l_1=2.5$ m, $E_2=1.2$ V To find: Balancing length of the wire ($l_2$) Formula: $\dfrac{E_1}{E_2}=\dfrac{l_1}{l_2}$ Calculation: From formula, $$l_2=\dfrac{l_1\times E_2}{E_1}=\dfrac{2.5\times1.2}{1.5}=2\text{ m}$$ Ans: The balancing length of the wire for another cell is 2 m.

17. 17.↑ back to the question

    Given: $R_1=12\ \Omega$, $l_2=120$ cm, $R_2=18\ \Omega$, $l'_2=150$ cm To find: Balancing length ($l_1$), Internal resistance (r) Formula: $r=R\left(\dfrac{l_1-l_2}{l_2}\right)$ Calculation: From first condition, $r=12\left(\dfrac{l_1-120}{120}\right)\qquad....(i)$ From second condition, $r=18\left(\dfrac{l_1-150}{150}\right)\qquad....(ii)$ From equation (i) and (ii), $$12\left(\dfrac{l_1-120}{120}\right)=18\left(\dfrac{l_1-150}{150}\right)$$ $\therefore 5(l_1-120)=6(l_1-150)$ $\therefore l_1=300$ cm Using equation (i) we get Internal resistance, $$r=12\left(\dfrac{l_1-120}{120}\right)=12\left(\dfrac{300-120}{120}\right)$$ $\therefore r=18\ \Omega$ Ans: i. The balancing length is 300 cm. ii. The internal resistance of the cell is 18 Ω.

18. 18.↑ back to the question

    Given: $l_1=3.2$ m, $l_2=0.7$ m To find: Ratio of emfs of two cells $\left(\dfrac{E_1}{E_2}\right)$ Formula: $\dfrac{E_1}{E_2}=\dfrac{l_1+l_2}{l_1-l_2}$ Calculation: From formula, $$\dfrac{E_1}{E_2}=\dfrac{3.2+0.7}{3.2-0.7}=\dfrac{3.9}{2.5}=1.56$$ Ans: The ratio of emfs of two cells is 1.56.