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A circuit shown in the figure has resistances `20 Omega` and `30 Omega`. At what value of resistance `R_x` will the thermal power generated in it be practically independent of small variations of that resistance? The voltage between points `A` and `B` is supposed to be constant in this case. |
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Answer» Let V be the potential difference applied across across A and B. Let `I,I_(1),I_(2)` be the currents through `R_(1),R_(2)` and `R_(x)` respectively. Here `R_(2)` and `R_(x)` are connected in parallel, their effective resistance is `R_(P) = (R_(2) R_(x))/(R_(2) + R_(x))` Total resistance between A and B is `R_(1) = (R_(2) R_(x))/(R_(2) + R_(x))` Current through the circuit, `I = V/(R_(1) + (R_(2) R_(x))/(R_(2) + R_(x)))` Potential difference across C and D `V_(G//D) I R_(P) = I = V/(R_(1) + (R_(2) R_(x))/(R_(2) + R_(x))) xx (R_(2) R_(x))/(R_(2) + R_(x))` `(VR_(2)R_(x))/(R_(1) (R_(2) + R_(x)) + R_(2)R_(x))` Current through resistance `R_(x)` will be `I_(2) = V_(CD)/R_(x) = (VR_(2))/(R_(1) (R_(2) + R_(x)) + R_(2)R_(x))` `= (VR_(2))/(R_(1) (R_(2) + R_(x)) + R_(2)R_(x))` Thermal power generated in resistance `R_(x)` will be `P = I_(2)^(2) R_(x) = (V^(2)R_(2)^(2)R_(x))/[R_(1)R_(2)+(R_(1) + R_(2))+R_(x)]^(2)` For P to be independent of `R_(x),dP/dR_(x) = 0` So, `d/dR_(x) [ (V^(2)R_(2)^(2)R_(x))/[R_(1)R_(2)+(R_(1) + R_(2))+R_(x)]^(2) ] = 0` On simplifying, we get `R_(1)R_(2) - (R_(1) + R_(2)) R_(x) = 0 or R_(x) = R_(1)R_(2)/(R_(1) + R_(2))` `:. R_(x) = (20 xx 30)/(20 + 30) = 12 Omega` |
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