A cell E 1 of emf 6   V and internal resistance 2 Ω is connected with another cell E 2 of emf 4…

A cell E1 of emf 6 V and internal resistance 2Ω is connected with another cell E2 of emf 4 V and internal resistance 8 Ω (as shown in the figure). The potential difference across points X and Y is:

  1. 10.0 V
  2. 5.6 V
  3. 2.0 V
  4. 3.6 V

Solution

I=6-410=15 A

 Vx+4+8×15-Vy=0

 Vx-Vy=-5.6⇒Vx-Vy=5.6 V

Asked in: JEE Main 2021 (24 Feb Shift 1)

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