At t = 4 sec
sources 30 u (t + 1) → exist and equals to 30 V 2 u (1 – t) → does not exist and equals to 0 amp Now, we will draw the equivalent circuit under this situation.

Correct Option: B

At t = 4 sec
sources 30 u (t + 1) → exist and equals to 30 V 2 u (1 – t) → does not exist and equals to 0 amp Now, we will draw the equivalent circuit under this situation.

At t = 0.5 sec
30 u (t + 1) → exist and equal to 30 V
2 u (1 – t) → exist and equal to 2 amp.

Now, we will draw the equivalent circuit under this situation we get this situation is same as the above question so,
I = 5 amp.

Correct Option: A

At t = 0.5 sec
30 u (t + 1) → exist and equal to 30 V
2 u (1 – t) → exist and equal to 2 amp.

Now, we will draw the equivalent circuit under this situation we get this situation is same as the above question so,
I = 5 amp.

At t = – 0.5 sec
source 30 u (t + 1) → exist and equal to 30V.
2 u (1 – t) → exist and equal to 2 amp.
Now, we will draw the equivalent circuit under this situation, we get

By applying source transformation theorem, we get

Now apply KVL to the loop, we get
30 + 20 – 10 I = 0

Correct Option: C

At t = – 0.5 sec
source 30 u (t + 1) → exist and equal to 30V.
2 u (1 – t) → exist and equal to 2 amp.
Now, we will draw the equivalent circuit under this situation, we get

By applying source transformation theorem, we get

Now apply KVL to the loop, we get
30 + 20 – 10 I = 0

At t = – 2 sec
source 30 u (t + 1) → does not exist and equal to 0 V 2 u (1 – t) → exist, and equal to 2 amp the equivalent circuit becomes

Now, apply superposition theorem, we conclude that only 20 V source is responsible for current I.

Correct Option: A

At t = – 2 sec
source 30 u (t + 1) → does not exist and equal to 0 V 2 u (1 – t) → exist, and equal to 2 amp the equivalent circuit becomes

Now, apply superposition theorem, we conclude that only 20 V source is responsible for current I.

Correct Option: C

=	1	x 3 x		12 sin		πt		6
	2					6

=	144 x 3			sin2		πt
	2					6