Answer:
Incomplete question
Check attachment for circuit diagram of the question, solution and explanation
Explanation:
a. Check attachment for the circuit diagram when the switch is at position "b" i.e. at t≥0
We want to find i(t) for t≥0
At this position "b", we will have an RL circuit which has an initial current when the switch was at position "a".
The current in this RL circuit is given as
i(t) = i(∞) + [i(0~) — i(∞)] exp(-t/τ)
τ is the time constant and it is given as
τ= L/R
Where,
L=200mH=200×10^-3=0.2H
R=2Ω
Then,
τ = L/R = 0.2/2
τ = 0.1 second
Then, we need to know the initial current and final current in the inductor.
i. At t=∞
ii. At t<0
i. At t=∞, we can still use the same circuit of switch at position " b", but, vat the time, there is no voltage drop at the inductor
Therefore, Using KVL for loop 1: sum of voltage in a loop is zero.
-24+2i=0
2i=24
Divide both sides by 2
i=24/2
i=12 A
The current at t=∞ is 12A
i(∞)= 12A
ii. Now at t<0, the switch is at position "a" for a long time.
Then, the inductor acts like a short circuit.
The circuit diagram is shown in attachment.
The current from the 8A source flows through the short circuit in opposite direction to i(0~)
Then, i(0~)=-8A
Now applying the formula
i(t) = i(∞) + [i(0~) — i(∞)] exp(-t/τ)
i(t) = 12 + [ —8 — 12] exp(-t/0.1)
i(t) = 12 — 20 exp(—10t) A
Then, the current in the inductor circuit at any time t is given as
i(t) = 12 — 20 exp(—10t) A.
b. Initial Voltage (Vo) i.e. at t=0
The voltage across an inductor is given as,
VL= Ldi/dt
Where L =0.2H
i(t) = 12 — 20 exp(—10t)
di(t)/dt = 200exp(—10t)
Therefore,
VL= Ldi/dt
VL=0.2×200exp(—10t)
VL = 40exp(—10t) Volts
Now, the initial voltage at t=0
VL=40exp(—10t)
VL(0)=40exp(—10(0))
Vo=40exp(0)
Vo = 40 Volts
The initial voltage in the inductor is 40 Volts
c. Time taken in milliseconds for inductor voltage to be 24V
t=?
From question b
VL = 40exp(—10t)
Now, we want to find t at VL=24V
VL = 40exp(—10t)
24 = 40exp(—10t).
Divide both sides by 40
24/40 = 40exp(—10t) / 40
0.6 = exp(—10t).
Take In of both sides
In(0.6) = In(exp(—10t))
-0.5108=-10t
Divide both sides by -10
t=-0.5108/-10
t=0.05108s to milliseconds
1000seconds= 1milliseconds
t = 51.08 milliseconds
t=51.08 ms
d. Yes, in the instant after the switch has been moved to position "b", the inductor sustains a current of 8A counterclockwise around the newly formed closed path. This current cause a drop of 16V across the 2ohms resistor. This voltage drops adds to the drop across the source, producing a 40V drop across the inductor.
e. Check attachment for graph of
i(t) vs t
v(t) vs t
Note : i(t) = 12 — 20 exp(—10t)
at t=0, i(0) =12-20= —8A
at t=∞, i(∞)=12-0=12A
Also, VL = 40exp(—10t)
At t = 0, v(0) = 40 V
At t = ∞ v(∞) = 0 V
