Tardigrade
Tardigrade - CET NEET JEE Exam App
Exams
Login
Signup
Tardigrade
Question
Physics
If current is decreasing at a rate of 1000 As -1, p . d between A and B is
Q. If current is decreasing at a rate of
1000
A
s
−
1
,
p
.
d
between
A
and
B
is
2153
251
Report Error
A
5V
33%
B
6V
67%
C
7V
0%
D
8V
0%
Solution:
V
A
−
(
2
×
2
)
−
3
−
1
0
−
3
×
(
−
1000
)
=
V
B
⇒
V
A
−
V
B
=
6
V