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 a weaker current than when any one of the batteries is connected to the same resistor. For this requirement to be fulfilled

Options

A. `(epsi_(2))/(epsi_(1))"must not lie between" (r_(2))/(r_(1)+R)and(r_(1))/(r_(2)+R)`
B. `(epsi_(2))/(epsi_(1))"must not lie between" (r_(2))/(r_(1)+R)and(r_(2)+R)/(r_(1))`
C. `(epsi_(2))/(epsi_(1))"must lie between" (r_(2))/(r_(1)+R)and(r_(1))/(r_(2)+R)`
D. `(epsi_(2))/(epsi_(1))"must not lie between" (r_(2))/(r_(1)+R)and(r_(2)+R)/(r_(1))`

Answer

B

Explanation

Let R' be the resistance of the heater of coil.
`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`