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Two batteries with e.m.f. 12 V and 13 V are connected in parallel across a load resistor ol 10 Omega . The internal resistances of the two batteries are 1 Omega and 2 Omega respectively. Tine voltage across the load lies between |
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Answer» `11.6 V and 11.7 V ` `E_(1) = 12 V, r_(1) = 1 Omega , R = 10 Omega` `E_(2) = 13 V, r_(2) = 2 Omega`, `I = ((E_(2))/(r_(1)) + (E_(2))/(r_(2)) )/(1 + (R)/(r_(1)) + (R)/(r_(2)))= ((12)/(5) + (13)/(2))/(1 + (10)/(1) + (10)/(2))` `therefore I = (12 + 6 - 5)/(1 + 10 + 5) = (18 - 5)/(16)` `therefore I = 1.156 ` A `therefore V = IR ` = `1.156 xx 10` `= 11.56 ` V ` therefore ` THUS it is between V = 11.5 V and V = 11.6 V |
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