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A power supply of fixed emf E is connect...

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



A

Graph I

B

Graph II

C

Graph III

D

GraphIV

Text Solution

Verified by Experts

The correct Answer is:
A
`I=(epsilon)/(R+R_(A)+R )`

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