A 6 , V cell with 0.5 , Omega internal resist | vedclass

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(C) Let $i_1$ be the current through the $6 \, V$ cell and $i_2$ be the current through the $10 \, V$ cell. The total current through the $12 \, \Omeg

Vedclass · Accepted Answer

$2.87$

Vedclass · Answer

(C) Let $i_1$ be the current through the $6 \, V$ cell and $i_2$ be the current through the $10 \, V$ cell. The total current through the $12 \, \Omega$ resistor is $(i_1 + i_2)$.Applying Kirchhoff's Voltage Law $(KVL)$ to the loop containing both cells:$10 - i_2(1) + i_1(0.5) - 6 = 0$$0.5 i_1 - i_2 = -4$ --- $(i)$Applying $KVL$ to the loop containing the $10 \, V$ cell and the external resistor:$10 - i_2(1) - (i_1 + i_2)(12) = 0$$10 - i_2 - 12 i_1 - 12 i_2 = 0$$-12 i_1 - 13 i_2 = -10$ or $12 i_1 + 13 i_2 = 10$ --- (ii)From $(i)$, $i_1 = 2(i_2 - 4) = 2 i_2 - 8$.Substituting into (ii):$12(2 i_2 - 8) + 13 i_2 = 10$$24 i_2 - 96 + 13 i_2 = 10$$37 i_2 = 106$$i_2 = 106 / 37 \approx 2.8648 \, A \approx 2.87 \, A$.

$A$ $6 \, V$ cell with $0.5 \, \Omega$ internal resistance, a $10 \, V$ cell with $1 \, \Omega$ internal resistance, and a $12 \, \Omega$ external resistance are connected in parallel. The current (in ampere) through the $10 \, V$ cell is

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