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

Question

A battery having e.m.f. 5 volt and internal resistance 0.5 Ω is connected with a resistance of 4.5 Ω , then the voltage at the terminals of battery is: (A) 4.5 V (B) 4 V (C) 0 V (D) 2 V

Tardigrade Admin · Accepted Answer

Current taken from the cell i=(E/R+r) Current in external resistance i=(V/R) ∴ (V/R)=(E/R+r) (V/4.5)=(5/4.5+0.5) ⇒ V=4.5 volt

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

4.5 V

4 V

0 V

2 V

Current taken from the cell $i = \frac{E}{R + r}$
Current in external resistance $i = \frac{V}{R}$
$∴ \frac{V}{R} = \frac{E}{R + r}$
$\frac{V}{4.5} = \frac{5}{4.5 + 0.5}$
$\Rightarrow V = 4.5 v o lt$

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

4.5 V

4 V

0 V

2 V

Current taken from the cell i=R+rE​ Current in external resistance i=RV​ ∴RV​=R+rE​ 4.5V​=4.5+0.55​ ⇒V=4.5volt

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

Current taken from the cell $i = \frac{E}{R + r}$
Current in external resistance $i = \frac{V}{R}$
$∴ \frac{V}{R} = \frac{E}{R + r}$
$\frac{V}{4.5} = \frac{5}{4.5 + 0.5}$
$\Rightarrow V = 4.5 v o lt$