Question

A battery having e.m.f. $5$ volt and internal resistance $0.5 \, \Omega $ is connected with a resistance of $4.5 \, \Omega $ , then the voltage at the terminals of battery is:

• A. 4.5 V
• B. 4 V
• C. 0 V
• D. 2 V

Answer 1

Answer: (A)

Current taken from the cell $i=\frac{E}{R+r}$
Current in external resistance $i=\frac{V}{R}$
$\therefore \frac{V}{R}=\frac{E}{R+r}$
$\frac{V}{4.5}=\frac{5}{4.5+0.5} $
$\Rightarrow V=4.5 \,volt$