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A resistance network is connected to a battery as shown in the figure below. If the intrrnal resistance of the battery is 5omega, then the value of R (in omega) for maximum power delivered to the network is |
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Answer» 2  Now in loop ABCA, USING KVL we have, `- 3Ri_(1)-4R(i_(1)-i_(2))+ 2Ri_(2)= 0` `- 3Ri_(1)-4Ri_(i)+4Ri_(2) + 2Ri_(2)=0` `7i_(1)=6i_(2)` `rArr i_(1)=(6)/(7)i_(2)` Also,`i_(1)+ i_(2)=i` Hence, `((6)/(7)+1)i_(2)=i or i_(2)=(7)/(13)i` `therefore i_(1)=(6)/(7)i_(2)=(6)/(7)XX(7)/(13)i rArr i_(1)= (6)/(13)i` Now from loop ABDA (which includes cell ), we have, `3Ri_(1) +2Ri_(2) = V` (nearly because internal resistance in not taken in ACCOUNT) `3Rxx(6)/(13)i+2Rxx(7)/(13)i=iR_(ed)` `rArr R_(eq)= R ((18 +14)/(13))=(32)/(13)R` For maximum power delivered to circuit, (internal resistancenof source )= external resistance) `rArr 5 = (32R)/(13) or R=(13xx5)/(32)=203 Omega` or `R = 2 Omega` |
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