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QJEEM23Ph4996: JEE Main 2023 Physics - Class 12 JEEM23Ph4996 Solution: A cell of emf 90 V is connected across series combination of…

A cell of emf 90 V is connected across series combination of two resistors each of 100Ω resistance. A voltmeter of resistance 400Ω is used to measure the potential difference across each resistor. The reading of the voltmeter will be :

Multiple Choice Options

  • A 40 V
  • B 90 V
  • C 45 V
  • D 80 V

Step-by-Step Verified Solution

JEE Main 2023 (Online) 24th January Evening Shift Physics - Current Electricity Question 85 English Explanation

Req=400×100500+100=180Ωi=90180=12 A Reading =12×400×100500=40 volt 

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