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Question Number 153057 by DELETED last updated on 04/Sep/21

Answered by DELETED last updated on 04/Sep/21

$${\mathrm{i}}_1+{\mathrm{i}}_2={\mathrm{i}}_3.....\left(1\right)$$$$\mathrm{\Sigma }\mathrm{E}+\mathrm{\Sigma }\mathrm{i}.\mathrm{R}=0\mathrm{hk}\mathrm{kirchoff}\mathrm{II}$$$$\mathrm{loop}1\left(\mathrm{searah}\mathrm{jarum}\mathrm{jam}\right)$$$$-4+{4\mathrm{i}}_1+{4\mathrm{i}}_3=0$$$${\mathrm{i}}_1+{\mathrm{i}}_3=1.....\left(2\right)$$$$\mathrm{loop}2\left(\mathrm{berlawanan}\mathrm{jarum}\mathrm{jam}\right)$$$$-2+{2\mathrm{i}}_2+{4\mathrm{i}}_3=0$$$${\mathrm{i}}_2+{2\mathrm{i}}_3=1....\left(3\right)$$$$\left(1\right)\to \left(2\right){\mathrm{i}}_1+{\mathrm{i}}_1+{\mathrm{i}}_2=1$$$${2\mathrm{i}}_1+{\mathrm{i}}_2=1....\left(4\right)$$$$\left(1\right)\to \left(3\right){\mathrm{i}}_2+2({\mathrm{i}}_1+{\mathrm{i}}_2)=1$$$${2\mathrm{i}}_1+{3\mathrm{i}}_2=1....\left(5\right)$$$$\left(4\right)\to \left(5\right)$$$${2\mathrm{i}}_1+{3\mathrm{i}}_2=1$$$${2\mathrm{i}}_1+{\mathrm{i}}_2=1$$$$-------$$$${2\mathrm{i}}_2=0\to {\mathrm{i}}_2=0$$$${2\mathrm{i}}_1+0=1\to {2\mathrm{i}}_1=1$$$${\mathrm{i}}_1=\frac{1}{2}\to {\mathrm{i}}_3=\frac{1}{2}+0=\frac{1}{2}$$$${\mathrm{Daya}}_{4\mathrm{\Omega }}={\mathrm{i}}^2\times \mathrm{R}={\left(\frac{1}{2}\right)}^2\times 4$$$$=\frac{1}{4}\times 4=1\mathrm{A}//$$

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