JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Application of Potentiometer

(1) To determine the internal resistance of a primary cell

(i) Initially in secondary circuit key K' remains open and balancing length \[({{l}_{1}})\] is obtained. Since cell E is in open circuit so it's emf balances on length \[{{l}_{1}}\] i.e. \[E=x{{l}_{1}}\]                        ...(i)

(ii) Now key K' is closed so cell E comes in closed circuit. If the process of balancing repeated again then potential difference V balances on length \[{{l}_{2}}\] i.e. \[E=x{{l}_{2}}\]                 ... (ii)

(iii) By using formula internal resistance \[r=\left( \frac{E}{V}-1 \right)\,.\,R'\] \[r=\left( \frac{{{l}_{1}}-{{l}_{2}}}{{{l}_{2}}} \right)\,.\,R'\]            

(2) Comparison of emf's of two cell : Let \[{{l}_{1}}\] and \[{{l}_{2}}\] be the balancing lengths with the cells \[{{E}_{1}}\] and \[{{E}_{2}}\] respectively then \[{{E}_{1}}=x{{l}_{1}}\] and \[{{E}_{2}}=x{{l}_{2}}\Rightarrow \frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\]

Let \[{{E}_{1}}>{{E}_{2}}\] and both are connected in series. If balancing length is \[{{l}_{1}}\] when cell assist each other and it is \[{{l}_{2}}\] when they oppose each other as shown then :

\[({{E}_{1}}+{{E}_{2}})=x{{l}_{1}}\]                          \[({{E}_{1}}-{{E}_{2}})=x{{l}_{2}}\]

\[\Rightarrow \]                               \[\frac{{{E}_{1}}+{{E}_{2}}}{{{E}_{1}}-{{E}_{2}}}=\frac{{{l}_{1}}}{{{l}_{2}}}\]         

or           \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{l}_{1}}+{{l}_{2}}}{{{l}_{1}}-{{l}_{2}}}\]   

(3) Comparison of resistances : Let the balancing length for resistance \[{{R}_{1}}\] (when XY is connected) is \[{{l}_{1}}\] and let balancing length for resistance \[{{R}_{1}}+{{R}_{2}}\] (when YZ is connected) is \[{{l}_{2}}\].

Then  \[i{{R}_{1}}=x{{l}_{1}}\] and \[i({{R}_{1}}+{{R}_{2}})=x{{l}_{2}}\Rightarrow \frac{{{R}_{2}}}{{{R}_{1}}}=\frac{{{l}_{2}}-{{l}_{1}}}{{{l}_{1}}}\]

(4) To determine thermo emf

(i) The value of thermo-emf in a thermocouple for ordinary temperature difference is very low (\[{{10}^{-6}}\] volt). For this the potential gradient \[x\] must be also very low \[({{10}^{-4}}\,\,V/m)\]. Hence a high resistance (R) is connected in series with the potentiometer wire in order to reduce current.

(ii) The potential difference across R must be equal to the emf of standard cell i.e. \[iR={{E}_{0}}\] \[\therefore \] \[i=\frac{{{E}_{0}}}{R}\]

(iii) The small thermo emf produced in the thermocouple \[e=xl\]

(iv) \[x=i\rho =\frac{iR}{L}\]\[\therefore \]\[e=\frac{iRl}{L}\]where L = length of potentiometer wire, \[\rho =\]resistance per unit length, \[l=\]balancing length for e 

(5) Calibration of ammeter : Checking the correctness of ammeter readings with the help of potentiometer is called calibration of ammeter.

(i) In the process of calibration of an ammeter the current flowing in a circuit is measured by an ammeter and the same current is also measured with the help of potentiometer. By comparing both the values, the errors in the ammeter readings are determined.

(ii) For the calibration of an ammeter, \[1\,\,\Omega \] standard resistance coil is specifically used in the secondary circuit of the potentiometer, because the potential difference across \[1\,\,\Omega \] is equal to the current flowing through it i.e. \[V=i\].

(iii) If the balancing length for the emf \[{{E}_{0}}\] is \[{{l}_{0}}\] then \[{{E}_{0}}=x{{l}_{0}}\Rightarrow x=\frac{{{E}_{0}}}{{{l}_{0}}}\] (Process of standardisation)

(iv) Let \[i\,'\] current flows through \[1\,\,\,\Omega \] resistance giving potential difference as \[V'=i'(1)=x{{l}_{1}}\] where \[{{l}_{1}}\] is the balancing length. So error can be found as  \[\Delta i=i-i'=i-x{{l}_{1}}=i-\frac{{{E}_{0}}}{{{l}_{0}}}\times {{l}_{1}}\]

(6) Calibration of voltmeter

(i) Practical voltmeters are not ideal, because these do not have infinite resistance. The error of such practical voltmeter can be found by comparing the voltmeter reading with calculated value of p.d. by potentiometer.

(ii) If \[{{l}_{0}}\] is balancing length for \[{{E}_{0}}\] the emf of standard cell by connecting 1 and 2 of bi-directional key, then \[x={{E}_{0}}/{{l}_{0}}\]. (iii) The balancing length \[{{l}_{1}}\] for unknown potential difference \[V'\] is given by (by closing 2 and 3)  \[V'=x{{l}_{1}}=({{E}_{0}}/{{l}_{0}}){{l}_{1}}\].

If the voltmeter reading is V then the error will be \[(V-V')\] which may be \[+ve,\,\,-ve\] or zero.