Q 12-03-008NEETNEET 2024Top questionEasy
The terminal voltage of the battery, whose emf is $10$ V and internal resistance $1\ \Omega$, when connected through an external resistance of $4\ \Omega$ as shown in the figure is :
Answer: (A) $8$ V
$$I = \frac{E}{R + r} = \frac{10}{4 + 1} = 2\ \text{A}$$
$$V = E - Ir = 10 - 2 \times 1 = 8\ \text{V}$$
Solution by Sreeraj P, M.Sc Physics