Question 81 of 380

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given:
${R}_{1}=15\Omega ,{R}_{2}=10\Omega ,{R}_{3}=20\Omega ,{R}_{4}=5\Omega ,{R}_{5}=25\Omega ,{R}_{6}=30\Omega ,E=15V$

A$9/32$B$7/18$C$13/24$D$20/3$

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Answer: (A) $9/32$

The equivalent resistance${R}_{eq}$ of this given network is,
${R}_{eq}=15+\dfrac{25}{3}+30=\dfrac{45+25+90}{3}=\dfrac{160}{3}$
By applying ohm's law, the Current $i$ through the battery is
$i=\dfrac{15}{\dfrac{160}{3}}=\dfrac{15\times 3}{160}$
$∴i=\dfrac{9}{32}A$