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In the circuit shown in figure, the emfs...

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?

Text Solution

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The correct Answer is:
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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