Saved Bookmarks
| 1. |
(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistance 0.015 Omegaarejoined in series to provide a supply to a resistance of 8.5 Omega . What are the current drawn fromthe supply and its terminal voltage ? (b) A secondary cell after long use has an emf of 1.9 V and a large internal resistance of 380 Omega. What maximum current can be drawn from the cell ? Could the cell drive the starting motorof a car? |
|
Answer» Solution :(a) Here emf of each cell `EPSI`= 2.0 V, INTERNAL resistance of each cell r = 0.015 `Omega `Number of cells n= 6 and external resistance R = 8.5 `Omega` ` therefore`In series arrangement total emf `epsi_(EQ) = n epsi = 6 xx 2.0 = 12 V ` and total internal resistance` r_(eq) = nr = 6 x 0.015 = 0.09 Omega` Current drawn from the SUPPLY`I = (epsi_(eq))/(R + r_(eq)) = (12 )/(8.5 + 0.09) = (12)/(8.59) = 1.4 A` `therefore ` Terminal voltage `V = epsi_(eq) - Ir_(eq) = 12 - 1.4 xx (0.09) = 11.9 V` (b) Here emf `epsi.` = 1.9 V, internal resistance `r. = 380 Omega`Maximum current which may be drawn from the cell `I. = (epsi.)/(r.) = (1.9)/(380) = 0.005A` As this current is extremely small, it cannot be used to drive the starting motor of a car. |
|