In a circuit shown in Fig, resistances `R_(1)` and `R_(2)` are known , as well as emf's `xi_(1)` and `xi_(2)`. The internal resitances of the sources are negatible. At what value of the resitance `R` will teh thermal power generated in it be the highest ? What is it equal to ?

Explanation

Indicate the currents in the circuit as shown in the fig.
Appying loop rule in the closed loop `12561, -Delta varphi = 0` we get
`i_(1) R - xi_(1) + iR_(1) = 0` ....(1)
and in the loop `23452`, .....(2)
`(i - i_(1)) R_(2) + xi_(2) - i_(1) R = 0` .....(2)
Solving (1) and (2) we get,
`i_(1) = (xi_(1) R_(2) + xi_(2) R_(1))/(R R_(1) + R_(1) R_(2) + R R_(2))`
So, thermal power genereated in the resistance `R`.
`P = i_(1)^(2) R = [(xi_(1) R_(2) + xi_(2) R_(1))/(R R_(1) + R_(1) R_(2) + R R_(2))]^(2) R`
For `P` to be maximum, `(dP)/(dR) = 0`, which fields
`R = (R_(1) R_(2))/(R_(1) + R_(2))`
Hence, `P_(max) = ((xi_(1) R_(2) + xi_(2) R_(1))^(2))/(4 R_(1) R_(2) (R_(1) + R_(2)))`