Current Electricity Ques 50 — JEE PYQ Chapterwise

  1. 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 $ $V$

(2019 Main, 8 April II)

(a) $13 / 24$

(b) $7 / 18$

(c) $20 / 3$

(d) $9 / 32$

Show Answer

Answer:

Correct Answer: 50.(d)

Solution:

Formula:

Combination of Resistances:

  1. Given circuit is redrawn and can be simplified as

So, current drawn through the cell is

$ \begin{aligned} i & =\frac{\text { Voltage }}{\text { Net resistance of the circuit }} \\ & =\frac{V}{R _{\text {eq }}^{\prime \prime}}=\frac{15}{(160 / 3)}=\frac{15}{160/3}=\frac{45}{160}=\frac{9}{32} A \end{aligned} $

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