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Two batteries of emf e_(1) and e_(2) having internal resistance r_(1) and r_(2) respectively are connected in series to an external resistance R. Both the batteries are getting discharged. The above described combination of these two batteries has to produce aweaker current than when any one of the batteries is connected to the same resistor. For this requirement to be fulfilled |
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Answer» <P>`(epsi_(2))/(epsi_(1))"must not lie between" (r_(2))/(r_(1)+R)and(r_(1))/(r_(2)+R)` `R'=(V^(2))/(P)=((100)^(2))/(1000)gt10Omega` If the heater has to operature with a power P'=62.5W, the voltage V' across its coil should be `V'=(P'R')^(1//2)=(6.25xx10)^(1//2)=25VA` Thus, out of 100V, a valtage drop of 25V occurs across the heater and the rest 100-25=75V occurs across the `10Omega` resistor. Therefore CURRENT in the cicruit is `I=(75)/(10)=7.5A` Now, current throght the heater =`(V')/(R)=(25)/(10)=2.5A` Therefore, current throgh reistor R=7.5-2.5=5.0A `"Here",R=(V')/(5.0A)=(25V)/(5.0)=5Omega` |
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