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For the circuit shown, with R₁=1.0 Ω, R₂=2.0 Ω, E₁=2 V and E₂=E₃=4 V, the potential difference between the points a and b is approximately (in V) [Figure: point a on the top wire, point b on the bottom wire. Left branch between a and b: a, resistor R₁ along the top to the left side, cell E₁ down the left side, resistor R₁ along the bottom to b. Middle branch: a, resistor R₂, cell E₂, b. Right branch: a, resistor R₁ along the top to the right side, cell E₃ and resistor R₁ down the right side, bottom wire to b. All three cells have their positive terminals toward the top wire.]

Asked in JEE Main 8th April 1st Shift 2019 · Kirchhoff's laws in networks

• (1) 2.7
• (2) 3.7
• (3) 2.3
• (4) 3.3

Step-by-step solution

Why the other options are wrong

• (1) Drops the E₁ branch from the numerator: (2+2)/(1.5)=2.7 V.
• (2) No standard slip gives 3.7 V.
• (3) No standard slip gives 2.3 V; reversing E₁ gives 2 V instead.

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