Question

. The switch has been in position a for a long time. At t = 0, the switch moves from position a to position b. The switch is a make-before-break type so there is no interruption of the inductor current. a. Find the expression for i(t) for t ≥ 0. b. What is the initial voltage across the inductor after the switch has been moved to position b? c. Does this initial voltage make sense in terms of circuit behavior? d. How many milliseconds after the switch has been put in position b does the inductor voltage equal 24V? e. Plot both i(t) and v(t) versus

Answer 1

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