Current Electricity 2 Question 3
3. 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) $9 / 32$
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Solution:
- Given circuit is redrawn and can be simplified as
So, current drawn through cell is
$$ \begin{aligned} i & =\frac{\text { Voltage }}{\text { Net resistance of the circuit }} \\ & =\frac{V}{R _{\text {eq }}^{\prime \prime}}=\frac{15}{(160 / 3)}=\frac{9}{32} A \end{aligned} $$
