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Question Number 121229 by sarahvalencia last updated on 06/Nov/20

Answered by som(math1967) last updated on 06/Nov/20

$$\mathrm{Total}\mathrm{resistance}\mathrm{of}\mathrm{the}\mathrm{circuite}$$$$8+\{1\mathbf{÷}(\frac{1}{10}+\frac{1}{15}+\frac{1}{20})\}$$$$=8+\frac{60}{13}$$$$=\frac{164}{13}\mathrm{ohm}$$$$\mathrm{current}\mathrm{flow}\mathrm{in}8\mathrm{ohm}$$$$48\times \frac{13}{164}=\frac{12\times 13}{41}\mathrm{amp}$$$$\mathrm{p}.\mathrm{d}\mathrm{accross}10\mathrm{ohm}$$$$(48-8\times \frac{12\times 13}{41})\mathrm{v}$$$$\mathrm{current}\mathrm{flow}10\mathrm{ohm}$$$$(48-\frac{96\times 13}{41})\mathbf{÷}10=1.75\mathrm{Amp}$$$$\left(\mathrm{approx}\right)$$$$$$

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