Describe briefly, with the help of a circuit diagram, how a potentiometer is used to determine the internal resistance of a cell.
Open in App
Solution
Measurement of internal resistance ofa cell using potentiometer is shown in figure. The cell of emf, E is connected across a resistance box R through key K2.
When K2 is open, balance length is obtained at length AN1=l1 E∝l1 E=ϕl1 ........ (i) When K2 is closed : Let V be the terminal potential difference of cell and the balance is obtained at AN2=l2 V∝l2 V=ϕl2 ........ (ii) From equations (i) and (ii), we get EV=l1l2 ..... (iii) E=I(r+R) V=IR EV=r+RR ....... (iv) From equations (iii) and (iv), we get R+rR=l1l2 ∴r=R(l1l2−1) We know l1,l2 and R, so we can calculate r.