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QJEEM19Ph2e67: JEE Main 2019 Physics - Class 12 JEEM19Ph2e67 Solution: In the figure shown, what is the current (in Ampere) drawn from …

In the figure shown, what is the current (in Ampere) drawn from the battery ?
You are given: R1 = 15 Ω\Omega , R2 = 10 Ω\Omega , R3 = 20 Ω\Omega , R4 = 5 Ω\Omega , R5 = 25 Ω\Omega , R6 = 30 Ω\Omega , E = 15 V JEE Main 2019 (Online) 8th April Evening Slot Physics - Current Electricity Question 245 English

Multiple Choice Options

  • A 13/24
  • B `7/18
  • C `9/32
  • D `20/3

Step-by-Step Verified Solution

Equivalent resistance of the given circuit

Req=45+10×5010+50=(45+506)Ω{R_{eq}} = 45 + {{10 \times 50} \over {10 + 50}} = \left( {45 + {{50} \over 6}} \right)\Omega

⇒\Rightarrow 1603Ω{{160} \over 3}\Omega

∴\therefore I=151603=932AI = {{15} \over {{{160} \over 3}}} = {9 \over {32}}A

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