JEE MAIN - Physics (2023 - 25th January Evening Shift - No. 23)
Two cells are connected between points A and B as shown. Cell 1 has emf of 12 V and internal resistance of 3$$\Omega$$. Cell 2 has emf of 6V and internal resistance of 6$$\Omega$$. An external resistor R of 4$$\Omega$$ is connected across A and B. The current flowing through R will be ____________ A.
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Answer
1
Explanation
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$\mathrm{KCL}$ at $A$ gives
$\frac{6-V_{A}}{4}+\frac{0-V_{A}}{6}+\frac{18-V_{A}}{3}=0$
$V_{A}=10$
So current through $4 \Omega=\frac{10-6}{4}=1 \mathrm{~A}$
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