Physics · 2021
JEE Main · 24 February 2021, Shift 1 · Q12
A cell E_1 of emf 6 V and internal resistance 2 Ω is connected with another cell E_2 of emf 4 V and internal resistance 8 Ω (as shown in the figure).…
A cell $\displaystyle \mathrm{E}_1$ of emf $\displaystyle 6$ V and internal resistance $\displaystyle 2 \Omega$ is connected with another cell $\displaystyle \mathrm{E}_2$ of emf $\displaystyle 4$ V and internal resistance $\displaystyle 8 \Omega$ (as shown in the figure). The potential difference across points X and Y is :
Official answer
(3)
5.$\displaystyle 6$ V
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JEE Main 2021 Physics question, with the answer from NTA’s final answer key. Where our answers come from.