The equivalent resistance for the ckt | Assessment

Solution 1
Part (a)
At t >>0 , the inductor behaves as a short circuit. So, the equivalent circuit is
The equivalent resistance for the ckt, Req= 2k + (12k||6k||4k) = 2k + 2k = 4kΩ
i= 10
4 k =2.5 mA
By current divider rule,
(i−i L)¿
iL= i(12k ∨¿ 6 k )
4 k +(12 k ∨¿ 6 k )=2.5 m∗4 k
8 k =1.25mA
Part (b)
EL=1
2 L iL
2 = 1
2∗16 m∗1.25 m∗1.25 m=12.5 nJ
Part (c)
At t > 0, the ckt is in transient state.
Time constant, τ = L
Rthev
Rthev =2 k ||12 k||6 k+ 4 k= 4
3 k + 4 k =16
3 kΩ

∴ τ= 16 m
16 k /3 =3 μs
Part (d)
At t = 0+, iL ¿
iL ( t ) =iL ( t ≫ 0 ) +¿ ¿
¿ 1.25 m+(0−1.25 m)e−t /3 μ
iL ( t ) =1.25 m(1−e−t /3 μ )
Part (e)
At t > 0,
Voltage across 6kΩ = Voltage across 4 kΩ + Voltage across L
 V 6 k=iL ( t )*4k + L d iL ( t )
dt
 V 6 k=1.25m(1−e−t /3 μ)*4k + 16m*1.25m* 1
3 μ ∗e−t /3 μ
¿ 5 (1−e
−t
3 μ )+ 20
3 e−t /3 μ
¿ 5 (1+ 1
3 e
−t
3 μ
)volt
Solution 2
Part (a)
At t < 0, it is assumed that the switch is closed for a long time so, capacitor
behaves as an open circuit i.e. iC ( t )=0, t < 0
 V C ( t ) =Voltage across 8 Ω=5V
At t >0, the circuit is redrawn as
To obtain Rthev, current source is open circuited,

Rthev = ( 2+10 ) ||6=12||6=4 Ω
so, time constant is τ =C Rthev =25 m∗4=100 ms = 1/10 s
∴V C ( t )=5 ( 1−e−10t ) volt , t>0
Part (b)
Calculating V C ( t ) for different time constants, we get the following table
T V C ( t )
τ 3.16 V
2τ 4.32 V
3τ 4.75 V
4
τ
4.90 V
5τ 4.966 V
6τ 4.987 V
7τ 4.995 V

The Plot: