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A battery of emf 2 V and initial resistance 1 Omega is connected across terminals A and B of the circuit shown in figure . |
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Answer» Thermal power generated in the EXTERNAL circuit will be maximum possible when `R = 16//25 Omega`. In the simplified circuit, the circuit is a balanced Wheatstone BRIDGE and a branch of `20//29 Omega and R ` is paralllel with this balanced bridge for maximum power.  `r = R_(external)` ` 1 = 1/(1/(3+6) + 1/(20//29+R)+1/(4+8))` or `R = 16/25 Omega` Maximum power developed in the external circuit is `P_(max) = i^2R = (2/(1+1))^2 xx 1 = 1W` Current through the upper branch `i_1 = i[((20/29+R)(4+8))/(9+(20/29 + R)+12)]` `i_2 = i[((20/29+R)(3+6))/(9+(20/29 + R)+12)]` Therefore, `i_1//i_2` is independent of R. |
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