Current Electricity Ques 50
- 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:
- 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} $
BETA
