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JEE Main 12 Apr 2019 Paper 1 Solved Paper
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© examsnet.com
Question : 8
Total: 90
To verify Ohm’slaw, a student connects the voltmeter across the battery as, shown in the figure. The measured voltage is plotted as a function of the current, and the following graph is obtained.
If
V
0
is almost zero, identify the correct statement.
The potential difference across the battery is 1.5 V when it sends a current of 1000 mA
The value of the resistance R is 1.5 Ω
The emf of the battery is 1.5 V and the value of R is 1.5 Ω
The emf of the battery is 1.5 V and its internal resistance is 1.5 Ω
Validate
Solution:
V
=
E
−
I
r
When
I
=
0
,
V
=
E
=
1.5
V
When
I
=
1000
m
A
,
V
=
0
∴
1.5
−
1
.
r
=
0
r
=
1.5
Ω
© examsnet.com
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