Saved Bookmarks
| 1. |
(a) The emf of a cell is E and its intenal resistance is r. Its terminals are connected to a resistance R. The potential difference between the terminal is 12 V if R = 3 Omega and 16 V if R = 8 Omega. Find the values of E and r (b) When an unknown resistance is connected across a series combination of two identical cells, each of 18 V, current through the resistance is 4 A. When it is connected across a parallel combination of the same cells, the current through it is 3 A. Find the unknown resistance and internal resistance of each cell. ( c) p identical cells, each of emf E and internal resistance r, are joined in series to form a closed circuit. One cell (X) is joined with reversed polarity. Find the potential difference across each cell except X and also across X. (d) 12 cells each having the same emf are connected in series and are kept in a closed box. Some of the cells are wrongly connected. This battery is connected in series with an ideal ammeter and two cells of same emf as of 12 cells. The current is 3 A when the cells and the battery aid each other and 2 A when the cells and battery oppose each other. How many cells in the battery are wrongly connected ? |
|
Answer» Solution :(a) `i = E//(r + R)` `p.d`. Across cell `V_A - V_B = V = (E - ir)` or `iR` `V = (ER)/(R + r)` When `R = 3 Omega, V = 12 V` `R = 8 Omega, V = 16 V` `12 = (E xx 3)/(3 + r)` …(i) `16 = (E xx 8)/(8 + r)`….(ii) (ii)//(i) `rArr (16)/(12) = (8)/(8 + r). (3 + r)/(3)` `(1)/(2) = (3 + r)/(8 + r) rArr 8 + r= 6 + 2r rArr r = 2 Omega` `12 = (3E)/(3 + r) = (3E)/(3 + 2) rArr E = 20 V` (b) Case (a) : `i_1 = (18 + 18)/(R + 2r) =(36)/(R + 2r)` `4 = (36)/(R + 2r) rArr R + 2r = 9` ...(i) Case (b) : Two cells of same EMF are equivalent to a cell of same emf `18 V` and resistance `r//2` `i_2 = (18)/(R + (r)/(2))` `3 = (18)/(R + (r)/(2)) rArr R + (r)/(2) = 6` ...(ii) Solving (i) and (ii), `r = 2 Omega, R = 5 Omega` ( C) One cell is wrongly connected Net emf `= (p -2) E` Net resistance `= pr` `i = ((p -2)E)/(pr)` All cells are supplying current except `X` `p.d` across each cell (except `X`) `V = E - ir = E - ((p -2)E)/(pr) r = E -((p -2)E)/(p)` =`(2E)/(p)` Cell `X` is TAKING current `p.d. V' = E + ir = E + ((p -2)E)/(pr) r = (2(p-1)E)/(p)`. (d) Let `n` cells are wrongly connected, let emf of each each cell is `E` and internal resistance is `r` New emf of battery `= (12 - 2n) E` Net resistance `= 12 r` (i) When battery and two cells aid each other Net emf `=(12 - 2n) E + 2 E` Net resistance `= 12 r+ 2r= 14 r` Current `i_1 = ((12 - 2n)E + 2E)/(14 r) = 3` ...(i) (ii) When battery and two cells oppose each other Net emf ` = (12 - 2n) E - 2 E` Current `i_2 = ((12 - 2n)E - 2E)/(14 r) = 2`...(ii) (i)//(ii) `rArr (14 - 2n)/(10 - 2n) = (3)/(2) rArr 28 - 4n = 30 - 6n` `2n = 2 rArr n = 1` only one cell is wrongly connected. (a)  (b)  
|
|