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Two cells of emf 2V and 4V and internal resistance 1Omega and 2Omega respectively are connected in parallel so as to send the current in the same direction through an external resistance of 10Omega. Find the potential difference across 10Omega resistor. |
Answer» Solution : `r_(2) = 2Omega` Applying KVL to mesh `E_(1)ABE_(1) "Equation" (1) rArr` `1I_(1) + 10( I_(1) + I_(2)) = 2"" 1I_(1) + 10(0.75) =2` `1I_(1)10I_(2) = 2 to (1)""1I_(1) = -7.5` APplying KVL to mesh `E_(2)ABE_(2)""11I_(1) = -5.5` `2I_(2) + 10(I_(1) + I_(2)) = 4""I_(1) = -0.5A` `10I_(1) + 10I_(2) = 2"""Potential drop across " 10Omega` `11I_(1) + 10I_(2) = 2""V = (I_(1) + I_(2))R` `10I_(1) + 12I_(2) = 4"" = (-0.5 + 0.75)10` `X^(LY)` eqn (1) by 10 and eqn (2) by 11 `"" = 0.25 xx 10` `110I_(1) + 100I_(2) = 20 "" V = 2.5V` `11OI_(1) + 132I_(2) = 44 ""` `110I_(1) + 132I_(2) = 44` `((-)(-)(-))/(-32I_(2) = -24)` `I_(2) = (24)/(32)` `I_(2)-0.75A` |
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