Download PHYS645 Electronics for Scientists

PHYS645 Electronics for Scientists
Assignment 1
1. Using KVL and KCL or mesh analysis,
determine the current Ix for the circuit shown
in Fig.1.
Solution:
Mesh1: I1=2A
Mesh 2:
K 4 $ I1 C ( 4 C 2 ) $ I2 = 10
Figure 1, For problem 1-3
I2=3A
Ix=I2=3A
2. Using a Thévenin equivalent, determine the current through the 2 ohm resistor in the
circuit shown in Fig.1.
Solution:
Calculate Rth (the circuit without R=2 ohm, and resourse U=10 V):
Rth = 4 U
Then calculate Vth (the circuit without R=2 ohm):
Vth = Voc = 10 C 2 $ 4 = 18
Ix =
Vth
18
=
=3A
Rth C 2 4 C 2
3. Using a Norton equivalent, determine the current through the 2 ohm resistor in the
circuit shown in Fig.1
First Rn :
Rn = Rth = 4 U ( from problem 2 )
Then In
( short load calculate short circuit current )
2 A = I1 C Isc
K10
I1 =
= K2.5
4
we get In = Isc = 4.5 A
Now Ix :
Rn
4
Ix =
$ In =
$ 4.5 = 3 A
Rn C 2
4C2
4. Using KVL and KCL or mesh analysis, determine the voltage Vx across the resistor in
the circuit shown in Fig.2.
Mesh 1 :
I1 = 6 A
(1)
Mesh 2 :
K3 $ I1 C ( 3 C 6 ) $ I2 K 6 $ I3 = 18 ( 2 )
Mesh 3 :
(3 )
I3 = K3 A
( 1 ) and ( 3 ) put into (2 ) ,
we get 9 $ I2 = 18, I2 = 2 A
Vx = 6 $ ( I2 K I3 ) = 6 $ ( K2 K ( K3 ) ) = 30 V
5. Using a Thévenin equivalent, determine the voltage Vx across the 6 ohm resistor in the
circuit shown Fig.2.
Rth=3 ohm;
Vth=Voc=I1*3+18=(6+3)*3+18=45 V;
Vx=6*Vth/(Rth+6)=45*6/(6+3)=30 V
6.Using a Norton equivalent, determine the voltage Vx across the 6 ohm resistor in the
circuit shown Fig.2.
Rn=Rth=3 ohm;
In =Isc=6+I1+3=6+(18V/3 ohm)+3=15 A;
Vx=In*(Rn//6 ohm)=15*(Rn*6)/(Rn+6)=15*3*6/(3+6)=30 V
Figure 2 For problem 4-6