In the figure shown, what is the current (in Ampere) drawn from the battery? You are given $$R_1=15 \Omega, R_2=10 \Omega, R_3=20 \Omega, R_4=5 \Omega, R_5=25 \Omega, R_6=30 \Omega, E=15 V$$

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Current Electricity

In the figure shown, what is the current (in Ampere) drawn from the battery? You are given

R_{1} = 15Ω, R_{2} = 10Ω, R_{3} = 20Ω, R_{4} = 5Ω, R_{5} = 25Ω, R_{6} = 30Ω, E = 15 V

Answer: (C)

The six resistors reduce in stages exactly as the circuit diagram shows.

R_{1} = 15 Ω

sits in series at the battery's positive terminal and

R_{6} = 30 Ω

in series at the return. The inner network made of

R_{2} = 10 Ω

R_{3} = 20 Ω

R_{4} = 5 Ω

and

R_{5} = 25 Ω

, connected as drawn, collapses to a single resistance of

\frac{25}{3} Ω

. These three parts are then in series, so the resistance the battery sees is

R_{e q} = R_{1} + \frac{25}{3} + R_{6} = 15 + \frac{25}{3} + 30 = \frac{45 + 25 + 90}{3} = \frac{160}{3} Ω

. The current drawn from the cell is therefore

I = \frac{E}{R _{e q}} = \frac{15}{160/3} = \frac{15 \times 3}{160} = \frac{45}{160} = \frac{9}{32} A

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option