The value of current through the 5 Omega resi | vedclass

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(D) Let the currents be as shown in the circuit diagram. By applying Kirchhoff's Current Law $(KCL)$ at the junction,the current through the $4 \Omega

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$\frac{1}{23} 	ext{ A}$

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(D) Let the currents be as shown in the circuit diagram. By applying Kirchhoff's Current Law $(KCL)$ at the junction,the current through the $4 \Omega$ resistor is $i = i_1 + i_2$.Applying Kirchhoff's Voltage Law $(KVL)$ to the left loop:$5 - 4(i_1 + i_2) - 8i_1 = 0$$5 - 4i_2 - 12i_1 = 0 \Rightarrow 12i_1 + 4i_2 = 5$ ... $(i)$Applying $KVL$ to the right loop:$8i_1 - 5i_2 - 3 = 0 \Rightarrow 8i_1 - 5i_2 = 3$ ... (ii)From $(i)$,$4i_2 = 5 - 12i_1 \Rightarrow i_2 = \frac{5 - 12i_1}{4}$.Substitute $i_2$ into (ii):$8i_1 - 5\left(\frac{5 - 12i_1}{4}ight) = 3$$32i_1 - 25 + 60i_1 = 12$$92i_1 = 37 \Rightarrow i_1 = \frac{37}{92} 	ext{ A}$.Now,find $i_2$:$i_2 = \frac{5 - 12(37/92)}{4} = \frac{5 - 37(3/23)}{4} = \frac{5 - 111/23}{4} = \frac{115 - 111}{4 	imes 23} = \frac{4}{4 	imes 23} = \frac{1}{23} 	ext{ A}$.Thus,the current through the $5 \Omega$ resistor is $\frac{1}{23} 	ext{ A}$.

The value of current through the $5 \Omega$ resistor of the given circuit is

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