A battery of emf E and internal resistance r is connected across a resistance R. Resistance R can be adjusted to any value greater than or equal to zero. A graph is plotted between the current passing through the resistance (I) and potential difference(V) across it. Select the correct alternatives

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

Emf of the battery is 10V

V-I graph can never be a straight lines as shown in figure

Internal resistance of the battery is 5 $Ω$

Maximum current that can be taken from the battery is 2A

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Potential difference across resistance is equal to the potential difference across the terminals of the battery. So V=E-Ir. This is an equation of a straight line. Comparing this with the given graph, we can see that E=10V and r=5 $Ω$ . Also
$I_{max} = \frac{E}{r} = \frac{10}{5} = 2 A$ When external resistance R=0.

Watch 3-min video & get full concept clarity