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(a) Six lead-acid type of secondary cells each of emf 2.0 V and internal resistacne 0.015Omega are joined in series to provide a supply to a resistance of 8.5 Omega. What are the current drawn from the suply and its terminal voltage ? (b) A secondary cell after long use has an emf of 1.9 Vand a large internal resistance of 380 Omega. What maximum current can be drawn from the cell? Could the cell drive tghe starting motor of a car ? |
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Answer» Solution :Here no. of rows, m = 1 No. of CELLS in one row, N = 6 emf of one cell, `epsilon ` =2 V External resistance , R = 8.5 `Omega` (a) Here, CURRENT passing above closed loop is, I = `("mne")/(mR + nr)` `therefore I = (1 xx 6 xx 2)/((1 xx 8.5 ) + (6 xx 0.015))` `therefore I = (12)/(8.59)` `therefore I = 1.3969 A = 1.4` A Now, terminal VOLTAGE of given source will be, V= IR = (1.4) (8.5) `therefore V = 11.9 ` Volt (b) Maximum current that can be drawn from given source is , `I_("max") = (epsilon)/(r) = (1.9)/(380)` ` therefore I_(max) = 0.005 ` A Motor of a car can not be started with above current because while starting the motor of a car (i.e. for triggering the engine of a car) minimum 100 A current is required. |
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