In the figure shown, what is the current (in Ampere) drawn from the battery? You are given: R1 = 15 Ω, R2 = 10 Ω, R3 = 20 Ω, R4 = 5 Ω, R5 = 25 Ω, R6 = 30 Ω, E = 15 V (a) 13/24 (b) 7/18 (c) 9/32 (d) 20/3

Answer: (c). R₃, R₄, R₅ are in series: 20 + 5 + 25 = 50 Ω, which is in parallel with R₂: (50 × 10)/(60) = 25/3 Ω. Req = R₁ + 25/3 + R₆ = 15 + 25/3 + 30 = 160/3 Ω I = (15)/(160/3) = 9/32 A. So the answer is (c).

In the figure shown, what is the current (in Ampere) drawn from the… | JEE Main 2019 (8 Apr, Shift 2)

Q65JEE Main 2019 · 8 APR · SHIFT 2Medium

Combination of Resistors

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

You are given: R1 = 15 $\Omega$ , R2 = 10 $\Omega$ , R3 = 20 $\Omega$ , R4 = 5 $\Omega$ , R5 = 25 $\Omega$ , R6 = 30 $\Omega$ , E = 15 V

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Answer: (c)

$R_3$ , $R_4$ , $R_5$ are in series: $20 + 5 + 25 = 50\ \Omega$ , which is in parallel with $R_2$ : $\dfrac{50 \times 10}{60} = \dfrac{25}{3}\ \Omega$ .
$R_{eq} = R_1 + \dfrac{25}{3} + R_6 = 15 + \dfrac{25}{3} + 30 = \dfrac{160}{3}\ \Omega$
$I = \dfrac{15}{160/3} = \dfrac{9}{32}$ A. So the answer is (c).

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