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The adjoining graph shows the variation of terminalpotential difference V, across a combination of three cells in series to a resistor, versus the current, I:(i) Calculate the emf of each cell. (ii) For what current I, will the power dissipation of the circuit be maximum ? |
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Answer» Solution : (i) We KNOW that V = `epsi - Ir` . From the graph it is CLEAR that when nocurrent is being drawn from the cells (i.e., I =0), voltage is 6.0 volt. As the battery is a combination of 3 cells and in an open circuitterminal potential difference is equal to emf, hence emf of each cell `epsi = (6.0)/(3) V = 2.0 V` (ii)From graph, terminal potential difference Vis zero when current drawn is `I_s=2.0 A` . It represents the short circuit CONDITION, where `0 = epsi - I_s. r ` and r is the internal resistance of the cell combination. ` rArr r = epsi/I_s = (6.0V)/(2.0A) = 3Omega` Power DISSIPATION of the circuit will be maximum when external resistance is equal to internal resistance (R=r) i.e., current drawn is `I = (epsi)/(r + R) = (epsi)/(2r) = (6V)/(2 xx 3Omega ) = 1A` |
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