Question

A battery emf 10 V and internal resistance 5 ohm is connected to a resistor.The current in the circuit is 0.5 A. What is the resistance of the resistor? What is the terminal voltage of the battery when the circuit is closed?

Answer 1

Given,

Emf of the batter = 10V  

Internal resistance = 5 ohm.

let R be the external resistance.

We know,  

current in the circuit, I =$\frac{E}{R+r}$

thus from above,

R = $\frac{E}{I}-r$

= $\frac{10V}{0.5 ohm}- 5ohm$

= 20 ohm -  5ohm

= 15 ohm.

Now,

terminal voltage , V = E -Ir

= 10V - (0.5A)(3 ohm)  

= 10V - 2.5 V  

= 7.5 V

Thus , terminal Voltage of the battery V is 8.5 V.

Answer 2

Explanation:

answer is 8.5 V

hope this helps you