Using Kirchhoff’s rules, calculate the current through the $40\Omega$ and $20\Omega$ resistors in the following circuit:
CBSE 55-1-2 PAPER SET 2019
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Apply $\ce{KVL}$ through $\text{ABCDA}$:
$80 - 20i_1 - 40(i_1 - i_2) = 0 80 - 60i_1 + 40i_2 = 0$ equation $(1)$ Apply $\text{KVL}$ through $\text{FEDCF}: 40 + 40(i_1 - i_2) - 10i_2 = 0 40 + 40i_1 - 50i_2 = 0$ equation $= 0 4i_2 - 6i_1 = -8$ equation $(1) -5i_2 + 4i_1 = -4$ equation $(2)$ Multiplying equation $(2)$ by $\frac{6}{4}$ and add with equation $(1)$
$\frac{-14}{4}\text{i}_2=-14,\text{i}_2=\frac{-14}{7}\times2$
$\text{i}_2=4\text{A}$ Put the value of $i_2$ in equation $(1) 4 \times (4) - 6i_1 = -8 16 - 6i_1 = -8 6i_1 = 16 + 8 = 24 i_1 = 4A$
So, current through $40\Omega$ resistor $= i_1 - i_2 = 4 - 4 = 0A$ Current through $20\Omega$ resistor $= 4A$
$80 - 20i_1 - 40(i_1 - i_2) = 0 80 - 60i_1 + 40i_2 = 0$ equation $(1)$ Apply $\text{KVL}$ through $\text{FEDCF}: 40 + 40(i_1 - i_2) - 10i_2 = 0 40 + 40i_1 - 50i_2 = 0$ equation $= 0 4i_2 - 6i_1 = -8$ equation $(1) -5i_2 + 4i_1 = -4$ equation $(2)$ Multiplying equation $(2)$ by $\frac{6}{4}$ and add with equation $(1)$
$\frac{-14}{4}\text{i}_2=-14,\text{i}_2=\frac{-14}{7}\times2$
$\text{i}_2=4\text{A}$ Put the value of $i_2$ in equation $(1) 4 \times (4) - 6i_1 = -8 16 - 6i_1 = -8 6i_1 = 16 + 8 = 24 i_1 = 4A$
So, current through $40\Omega$ resistor $= i_1 - i_2 = 4 - 4 = 0A$ Current through $20\Omega$ resistor $= 4A$
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