The terminal voltage of a battery, whose emf is $15 \mathrm{~V}$ and internal resistance is $1 \Omeg

Question

The terminal voltage of a battery, whose emf is $15 \mathrm{~V}$ and internal resistance is $1 \Omega$, when connected across an external resistance of $4 \Omega$ as shown in the figure is

Professor PC Thomas Classes · Answer

(d) : The current in the given circuit is $I = \frac{E}{R+r} = \frac{15}{4+1} = \frac{15}{5} = 3 \mathrm{~A}$. The terminal voltage across the battery is $V = E - Ir = 15 - (3 	imes 1) = 12 \mathrm{~V}$.