Q 12-03-263JEE MainJEE Main 2019 (8 Apr, Shift 2)Easy
In the figure shown, what is the current (in ampere) drawn from the battery? You are given: $R_1 = 15\ \Omega$, $R_2 = 10\ \Omega$, $R_3 = 20\ \Omega$, $R_4 = 5\ \Omega$, $R_5 = 25\ \Omega$, $R_6 = 30\ \Omega$, $E = 15\ \text{V}$
Answer: (A) $9/32$
$R_3$, $R_4$ and $R_5$ are in series: $20 + 5 + 25 = 50\ \Omega$. This is in parallel with $R_2$:
$$\frac{50\times10}{60} = \frac{25}{3}\ \Omega$$
Adding $R_1$ and $R_6$ in series: $R_{eq} = 15 + \frac{25}3 + 30 = \frac{160}{3}\ \Omega$.
$$I = \frac{15}{160/3} = \frac{45}{160} = \frac{9}{32}\ \text{A}$$
Solution by Sreeraj P, M.Sc Physics