Three resistors having resistances of 1.6 Ω, 2.4 Ω, 4.8 Ω are connected in parallel to a 28V battery that has negligible resistance. The current through the battery is

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Three resistors having resistances of $1.6 Ω, 2.4 Ω, 4.8 Ω$ are connected in parallel to a 28V battery that has negligible resistance. The current through the battery is

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40 A

35 A

30 A

45 A

answer is B.

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Detailed Solution

\frac{1}{R_{p}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}

$= \frac{1}{1.6} + \frac{1}{2.4} + \frac{1}{4.8}$

$= \frac{10}{16} + \frac{10}{24} + \frac{10}{48}$

$= \frac{10}{8} [\frac{1}{2} + \frac{1}{3} + \frac{1}{6}]$

$= \frac{5}{4} [\frac{3 + 2 + 1}{6}]$

$= \frac{5}{4} \times \frac{6}{6} = \frac{5}{4}$

$\frac{1}{R_{p}} = \frac{4}{5}$

$V = IR$

$I = \frac{V}{R_{p}} = \frac{28}{\frac{4}{5}} = 28 \times \frac{5}{4} \Rightarrow I = 35 A$

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