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QJEEM19Ph9db6: JEE Main 2019 Physics - Class 12 JEEM19Ph9db6 Solution: In the given circuit, an ideal voltmeter connected across the…

In the given circuit, an ideal voltmeter connected across the 10 Ω\Omega resistance reads 2V. The internal resistance r, of each cell is: JEE Main 2019 (Online) 10th April Morning Slot Physics - Current Electricity Question 235 English

Multiple Choice Options

  • A 1.5 Ω\Omega
  • B 1 Ω\Omega
  • C 0.5 Ω\Omega
  • D 0 Ω\Omega

Step-by-Step Verified Solution

For the given circuit

JEE Main 2019 (Online) 10th April Morning Slot Physics - Current Electricity Question 235 English Explanation Given that VAB = 2 V

∴\therefore I = 215+210=13A{2 \over {15}} + {2 \over {10}} = {1 \over 3}A

Also I(2r + 2) = 1.5 + 1.5 – VAB

⇒\Rightarrow 2r + 2 = (3–2)3

⇒\Rightarrow r=12Ωr = {1 \over 2}\Omega

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