In the figure shown, what is the current (in Ampere) drawn from the battery ? You are given $R_1 = 15\,\Omega $$,R _2 = 10\,\Omega ,$$ R_3 = 20\,\Omega ,$$ R_4 = 5\,\Omega ,$$R_5 = 25\,\Omega ,$$R_6 = 30\,\Omega , $$E = 15\,V$
JEE MAIN 2019, Medium
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$R_{e q}=15+\frac{25}{3}+30=\frac{45+25+90}{3}=\frac{160}{3}$
$\mathrm{I}=\frac{\mathrm{E}}{\mathrm{R}_{\mathrm{eq}}}=\frac{15 \times 3}{160}=\frac{9}{32} \,\mathrm{amp}$.
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