The terminal voltage of the battery, whose emf is $10 \mathrm{~V}$ and internal resistance $1 \Omega$, when connected through an external resistance of $4 \Omega$ as shown in the figure is:
A$6 \mathrm{~V}$
B$8 \mathrm{~V}$
C$10 \mathrm{~V}$
D$4 \mathrm{~V}$
NEET 2024, Medium
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B$8 \mathrm{~V}$
b $\text { Current in circuit } i =\frac{10}{4+1}=2 \mathrm{~A}$
$\text { Terminal voltage } =E-i R$
$=10-2 \times 1=8 \mathrm{~V}$
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