A cell E1 of emf 6V and internal resistance 2Ω is connected with another cell E2 of emf 4 V and internal resistance 8Ω (as shown in the figure). The potential difference across points X and Y is :
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a
10.0 V
b
3.6 V
c
5.6 V
d
2.0 V
answer is C.
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Detailed Solution
Current in the circuit, I=VnetRnet=6−48+2=15AUsing KVL, Vx+4+8×15=Vy ⇒Vy-Vx=5.6V