JEE PYQ: Current Electricity Question 41
Question 41 - 2019 (08 Apr 2019 Shift 2)
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.
(1) 13/24
(2) 7/18
(3) 9/32
(4) 20/3
Show Answer
Answer: (3)
Solution
$R_1$, $R_5$ and $R_6$ are in series: $R = 20 + 5 + 25 = 50,\Omega$. This is parallel with $R_2$: $\frac{10 \times 50}{10 + 50} + 15 + 30 = \frac{160}{3},\Omega$. $I = \frac{15}{160/3} = \frac{9}{32}$ A.
BETA
