Notes on Application of Potentiometer | Grade 12 > Physics > Electrical Circuits | KULLABS.COM

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#### Application of Potentiometer

A potentiometer basically measures the potential difference between two points.

##### Determination of Internal Resistance of a Cell

A cell of emf E whose internal resistance r is to be determined is connected in the potentiometer circuit. The positive terminal of E is connected to A where positive terminal of the driving cell E­0 is connected and the negative terminal to a galvanometer. A resistance box R is connected parallel to the cell through a key K. a steady current is passed through the wire by the driving cell.

Initiallly the key is open and the emf E of the cell is balanced in the potentiometer wire. Suppose the balanced point obtained at D and let l1 be the length of wire AD. Then E is balanced by p.d. VAD across AD. So,

\begin{align*} E &= V_{AD} \\ \text {From the principle of potentiometer} \\ V_{AD} &\propto l_1 \\ E &\propto l_1 \dots (i)\\ \end{align*}

Now, a known resistance R is provided by R.B. and the key K is closed. A current I’ will pass in the closed circuit of R and E. then a terminal p.d., V is obtained across the cell which is again balanced by the p.d. in the potentiometer wire. Let D’ be the point at which the null deflection is obtained and p.d. across AD’ be VAD’. So,

\begin{align*} V = V_{AD}’ \end{align*}If l2is the length of this portion AD’ of the wire, from the principle of potentiometer \begin{align*} V_{AD}’ \propto l_2 \\ \text {and} \: V \propto l_2 \dots (ii) \\ \text {Dividing equation} \: (i) \: \text {by equation} \: (ii) , \: \text {we get} \\ \frac EV &= \frac {l_1}{l_2} \\ \text {As} \: E &= I’ (R + r) \\\ \text {and the terminal p.d.,} \: V &= I’R \\ \end{align*}

\begin{align*} \text {Substituting these values in equation} \: (iii), \: \text {we get} \\\frac {I’(R +r)}{I’R} &= \frac {l_1}{l_2} \\ \text {or,} \: \frac {R + r}{R} &= \frac {l_1}{l_2} \\ \text {or,} \: 1 + \frac rR &= \frac {l_1}{l_2} \\\text {or,} \: \frac rR &= \frac {l_1}{l_2} – 1 \\ \text {or,} \: r &= R \left (\frac {l_1}{l_2} – 1\right ) \\ \text {or,} \: r &= R \left ( \frac {l_1 – l_2}{l_2} \right ) \dots (iv) \\ \end{align*}As l1, l2and R are known, the internal resistance r of the cell can be determined.

#### Comparison of emfs of two Cells and Determination of emf of a Cell

Two cells whose emfs E1 and E2 are to be compared are connected in the potentiometer circuit as shown in the figure. The driving cell of emf E0 maintains a steady current in the circuit of the potentiometer wire AB. The positive terminals of E1 and E2 are connected at A where positive terminal of E0 is connected and the negative terminals to the two-way key. A galvanometer is connected between the key and the jockey that slides over the wire. E0 must be greater than E1 and E2.

One of the cells say E1 is connected in the circuit by closing K1 of the two-way key and E2 disconnected by making K2 open. The jockey is slide along the wire AB to find the null point. When the jockey is placed at C near A on the wire, the length of portion AC is small and p.d., VAC in it is small. As the emf E1 is greater than VAC, the current will flow through the galvanometer, G in the direction of E1 and the galvanometer is deflected in the left direction.

When the jockey is placed at D near B, the p.d., VAD across portion AD will be greater than E1 and the current will flow through, G in opposite direction of E1. So, the galvanometer shows deflection in right direction. This work confirms that the circuit is correct and by trail and error method, a point says F is found at which G shows null deflection. At this condition, p.d. VAF across AF is equal to emf, E1 and in the galvanometer coil. So, whole emf E1 is balanced by the p.d. VAF in the length l1 of the wire. So, at the balanced condition,

\begin{align*} E_1 = V_{AF} \\ \text {From the principle of potentiometer,} \: V_{AF} \propto l_1 \\ \text {So,} \: E_1 \propto l_1 \dots (i) \\ \text {Similar work is repeated for the next cell,} \: E_2. \\ \text {Then we will get} \\ E_2 \propto l_2 \dots (ii) \\ \text {Dividing equation} \: (i) \: \text {by equation} \: (ii), \: \text {we get} \\ \frac {E_1}{E_2} &= \frac {_1}{l_2} \\ \text {Since} \: l_1 \: \text {and} \: l_2 \: \text {are measured in the potentiometer wire,} \\ \frac {E_1}{E_2} \: \text { can be determined.} \\\end{align*} If the emf of one cell, say E2is known, the emf of the other cell can be determined as $$E_1 = \frac {l_1}{l_2}E_2$$

reference

Manu Kumar Khatry, Manoj Kumar Thapa,et al. Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.

S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.

A potentiometer basically measures the potential difference between two points.

\begin{align*} \text {or,} \: r &= R \left ( \frac {l_1 – l_2}{l_2} \right ) \dots (iv) \\ \end{align*}As l1, l2and R are known, the internal resistance r of the cell can be determined.

If the emf of one cell, say E2is known, the emf of the other cell can be determined as $$E_1 = \frac {l_1}{l_2}E_2$$.

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