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Question Number 58203 by Umar last updated on 19/Apr/19
$$\mathrm{A}\:\mathrm{battery}\:\mathrm{can}\:\mathrm{supply}\:\mathrm{current}\:\mathrm{of}\:\mathrm{1}.\mathrm{2A}\: \\ $$$$\mathrm{and}\:\mathrm{0}.\mathrm{4A}\:\mathrm{through}\:\mathrm{4}\Omega\:\mathrm{and}\:\mathrm{14}\Omega\:\:\mathrm{respectively}. \\ $$$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{internal}\:\mathrm{resistance}\:\mathrm{of}\:\mathrm{the}\:\mathrm{battery} \\ $$
Answered by mr W last updated on 19/Apr/19
$${let}\:{X}={internal}\:{resistance}\:{of}\:{battery} \\ $$$$\mathrm{1}.\mathrm{2}×\left(\mathrm{4}+{X}\right)=\mathrm{0}.\mathrm{4}×\left(\mathrm{14}+{X}\right) \\ $$$$\Rightarrow{X}=\mathrm{1}\Omega \\ $$
Commented by Umar last updated on 19/Apr/19
$$\mathrm{thanks} \\ $$