Instrumentation: Circuit Analysis

INSTRUMENTATION 1
INSTRUMENTATION
By Name
Course
Instructor
Institution
Location
Date

INSTRUMENTATION 2
QUESTIONONE
Draw Laplace transform circuit of the given circuit,
S is the complex frequency;
As for t ¿ 0: 124 (t) Short circuit
And assume circuit work in steady state So ;
I(0) = I (0) =i(o) =0omp
For t¿ 0 ; 124 (t) = 12 Volt
And from ;

INSTRUMENTATION 3
Apply nodal analysis at node 1;
V 1
12 + V 1
2 + V 1−12
6 = 0
V1 [ V 1
12 + 1
2 + 1
6 ] =2
V1= 2.666 volt
I(∞) = 2.666
2 amp
I(∞ ¿ =1.333 amps
As we know that inductor current i(t) for t¿ 0 for RL circuit with source given by .
I(t) = (∞ ¿ + ( i(0)- (∞ ¿ e
−t
τ
τ is time constant = L
Req

INSTRUMENTATION 4
Req is calculated across inductor by deactivating voltage source by short circuit as shown below;
Therefore τ =3
2 11

INSTRUMENTATION 5
So: i(t) = 1.333+(0-1.333) e
−t
3
2
Ui(t) = 1.333((1- e
−2 t
3 ) amp for t¿ 0
As ; V3Ω = L d
dt i(t)
= 3* d
dt ( 1.333*((1-1.333) e
−2 t
3
= 3*1.333(-(- 2
3 ) e
−2 t
3
V3 Ω=2.666 e
−2 t
3 for t ¿ 0
As i3 Ω= V 3 Ω
3 amp
= 2.666
3 e
−2 t
3
i3 Ω = 0.006e
−2 t
3

INSTRUMENTATION 6
Apply KCL at node a
i2= i(t) +i3 Ω
= 1.333(1-e
−2 t
3 ) + 0.006 e
−2 t
3
So
V= 2* i2 Ω [ 1.333(1- e
−2 t
3 ) +0.006 e
−2 t
3 ]
V(t) = 2.666(1-e
−2 t
3 )+1.772 e
−3 t
2 for t¿ 0
Question 2
V0= 10+5e-50t
Vo(0) = 10+5 =15
At t=0
Vc= Ce - t
R 1 ×1 0−6 +A ( comparing this with V0(t)
R1= 1
50× 10−6 = 20kΩ