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A 10 V cells of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38Q . As shown in the figures , Find the value of current in the circuit In a potentiometer arrangement for determining the emf . of a cell the balance point of the cell in open circuit is 350 cm. When a resistance of 9Q . is used in the external circuit of the cells the balance point shift to 300 cm. Determine the internal resistance of the cell. |
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Answer» Solution :Writing the equation FINDING the current By Kirchoffs law , we have ,for the LOOP ABCD `+ 200 -38i -1 (0) =0 ` ` therefore I =( 190 )/( 38 ) A = 5A ` ` (##DBT_SM_PHY_XII_DL_18_E02_002_S01.png" width="80%"> Finding the Net emf Stating that `I= (V)/® ` Calculating The two cells being in opposition ` therefore ` net emf = ` ( 200 -10 ) V = 190 V ` Now `I= (V) /(R) ` [Note : some students may use the formule] ` ( in )/( r) =( in _ 1) /( r_1) +(in _2) /(r_2) ` and ` "" r=( r_1r_2)//( r_1+ r_2)` For two cells connected in parallel they may then say that ` r=0` ` in ` is indeterminate and hence I is also indeterminate Award full marks (2) to students giving This line of REASONING |
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