In the circuit shown S1 and S2 are switches which remains closed for a long time. Now S3 and S4 are closed simaltaneously . Just after S3 and S4 are closed, I1 and I2 are the currents as shown in circuit .
Answers
Answered by
0
Answer:
Correct option is
B
I
max
=
4R
V
D
τ=
R
2L
ℓn2
i
max
=(i
2
−i
1
)
max
Δi=(i
2
−i
1
)=
R
V
[1−e
−(
2L
R
)t
]
−
R
V
[1−e
(
L
R
)t
]
R
V
[e
−(
L
R
)t
−e
−(
2L
R
)t
]
For (Δi)
max
dt
d(Δi)
=0
R
V
[−
L
R
e
−(
L
R
)t
−(−
2L
R
)e
−(
2L
R
)t
]=0
e
−(
L
R
)t
=
2
1
e
−(
2L
R
)t
e
−(
2L
R
)t
=
2
1
(
2L
R
)t=ℓn2
t=
R
2L
ℓn2→ time when i is maximum.
i
max
=
R
V
[e
L
R
(
R
2L
ℓn2)
−e
−(
2L
R
)(
R
2L
ℓn2)
]
∣I
max
∣=
R
V
∣
∣
∣
∣
∣
[
4
1
−
2
1
]
∣
∣
∣
∣
∣
=
4
1
R
V
Answered by
0
Answer:
In the circuits as shown in the figure, S
1
and S
2
are switches. S
2
remains closed for a long time and S
1
is opened. Now S
1
is also closed. Just after S
1
is closed, find the potential difference (V) across R and
dt
di
in L.
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