A power supply of fixed emf E is connected to a variable resistor (resistance = R) and an ammeter of resistance `R_(A)`. The internal resistance of supply is r. The potential difference between the terminals of the power supply in closed circuits is denoted by V and current in circuit denoted by I. A graph is obtained between V and I by drawing varying amount of current from the supply. The correct `V - I` graph for the given situation is

Options

A. Graph I
B. Graph II
C. Graph III
D. GraphIV

Answer

A

Explanation

`I=(epsilon)/(R+R_(A)+R )`

`V+epsilon-(epsilon_(r ))/((R+R_(A)+r ))=epsilon-Ir = (epsilon(R+R_(A)))/((R+R_(A)+v))`