QJEEM19Ph81b4: JEE Main 2019 Physics - Class 12 JEEM19Ph81b4 Solution: For the circuit shown, with R 1 = 1.0W, R 2 = 2.0 W, E 1 = 2 V…

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QJEEM19Ph81b4: JEE Main 2019 Physics - Class 12 JEEM19Ph81b4 Solution: For the circuit shown, with R 1 = 1.0W, R 2 = 2.0 W, E 1 = 2 V…

For the circuit shown, with R₁ = 1.0W, R₂ = 2.0 W, E₁ = 2 V and E₂ = E₃ = 4 V, the potential difference between the points 'a' and 'b' is approximately (in V): JEE Main 2019 (Online) 8th April Morning Slot Physics - Current Electricity Question 248 English

JEE Main 2019 (Online) 8th April Morning Slot Physics - Current Electricity Question 248 English

Multiple Choice Options

A 3.3
B 2.7
C 2.3
D 3.7

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Accepted Answer

Given circuit is

JEE Main 2019 (Online) 8th April Morning Slot Physics - Current Electricity Question 248 English Explanation 1

Above circuit can be viewed as

JEE Main 2019 (Online) 8th April Morning Slot Physics - Current Electricity Question 248 English Explanation 2

This is a parallel combination of three cells or in other words, a parallel grouping of three cells with internal resistances.

So, $V_{a b} = E_{e q} = \frac{I_{e q}}{r_{e q}} = \frac{\frac{E_{1}}{r_{1}} + \frac{E_{2}}{r_{2}} + \frac{E_{3}}{r_{3}}}{\frac{1}{r_{1}} + \frac{1}{r_{2}} + \frac{1}{r_{3}}}$

$= \frac{\frac{2}{2} + \frac{4}{2} + \frac{4}{2}}{\frac{1}{2} + \frac{1}{2} + \frac{1}{2}} = \frac{10}{3} V \approx 3.3 V$

QJEEM19Ph81b4: JEE Main 2019 Physics - Class 12 JEEM19Ph81b4 Solution: For the circuit shown, with R 1 = 1.0W, R 2 = 2.0 W, E 1 = 2 V…

Multiple Choice Options

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