In the circuit shown in figure, the emfs of batteries are `E_1`, and `E_2` which have internal resistances `R_1`, and `R_2`. At what value of the resistance `R` will the thermal power generated in it be the highest? What it is?

In the circuit shown in figure, the emfs of batteries are `E_1`, and `

1.	In the circuit shown in figure, the emfs of batteries are `E_1`, and `E_2` which have internal resistances `R_1`, and `R_2`. At what value of the resistance `R` will the thermal power generated in it be the highest? What it is?
Answer» Correct Answer - A::B::D The two batteries are in parallel.. Thermal power generated in `R` will be miximum when, total internal resitance = total external resitance `or R=(R_1R_2)/(R_1+R_2)` `E_(eq)=((E_1/R_1+E_2/R_2))/((1/R_1+1/R_2))` `=((E_1R_2+E_2R_1)/(R_1+R_2))` `R_("net")=(2R_1R_2)/(R_1+R_2)` `:. i=E_(eq)/R_("net")=(E_1R-2+E_2R_1)/(2R_1R_2)` Maximum power through R `P_(max)=i^2R=((E_1R_2+E_2R_1)^2)/(4R_1R_2(R_1+R_2))`

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In the circuit shown in figure, the emfs of batteries are `E_1`, and `E_2` which have internal resistances `R_1`, and `R_2`. At what value of the resistance `R` will the thermal power generated in it be the highest? What it is?

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