solutions of homework problems
DESCRIPTION
Solutions of Homework problems. Resistive circuits. Problem 1 Use KVL and Ohms law to compute voltages v a and v b . -. v 1. From Ohms law: v 1 =8k W* i 1 =8[V]. v 2 =2k W* i 2 =-2[V] Form KVL: v a =5[V]-v 2 =7[V] v b =15[V]-v 1 -v a =0[V]. +. -. v 2. +. +. +. -. -. - PowerPoint PPT PresentationTRANSCRIPT
Solutions of Homework problems
Resistive circuits
Problem 1Use KVL and Ohms law to compute voltages va and vb .
+
++
+
v2-
--
-v1
From Ohms law:v1=8kW*i1=8[V]v2=2kW*i2=-2[V]
Form KVL:va=5[V]-v2=7[V]vb=15[V]-v1-va=0[V]
Resistive circuits
Problem 2Write equations to compute voltages v1 and v2 , next find the current value of i1
From KCL:50 mA=v1/40+(v1-v2)/40 and100 mA=v2/80+(v2-v1)/40
Multiply first equation by 40:2=v1+v1-v2=2v1-v2
From second equation:8=v2+2(v2-v1)=3v2-2v1 add both sides:10=2v2 => v2=5 [V], v1=1+v2 /2=3.5[V] i1= (v1-v2)/40=-1.5/40=37.5 [mA]
50 mA
W40
W40 W80 100 mA
i1i1v2v1
Thevenin & Norton
Problem 3: Find Thevenin and Norton equivalent circuit for the network shown.
I1 N2
I2 vt
N1
From KVL
Thevenin & Norton
I1 N2
I2Isc
N1
From KVL
Thevenin & Norton
Note: Negative vt indicates that the polarity is reversed and as a result this circuit has a negative resistance.
+_Vt=-6 V
RTh=-1.33ΩA
B
Thevenin Equivalent
In=4.5 A RTh=-1.33Ω
A
B
Norton Equivalent
RTh=vt/Isc=-1.33Ω
Problem 4: Find the current i and the voltage v across LED diode in the circuit shown on Fig. a) assuming that the diode characteristic is shown on Fig. b).
Draw load line. Intersection of load line and diode characteristic is the i and v across LED diode: v ≈ 1.02 V and i ≈ 7.5 mA.
Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.
-10 -5 0 5 100
1
2
3
4
5
v (V)
i (m
A)
(a)
Diode is on for v > 0 and R=2kΩ.
+
v_
2kΩ
i
In a series connection voltages are added for each constant current
Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.
-10 -5 0 5 100
1
2
3
4
5
v (V)
i (m
A)
(a) +
v_
2kΩ
i
Resulting characteristics
Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.
-10 -5 0 5 100
1
2
3
4
5
v (V)
i (m
A)
(b)Due to the presence of the 5V supply the diode conducts only for v > 5, R = 1kΩ
+
v_
1kΩi
+_
5V
First combine diode and resistance then add the voltage source
(c)
-10 -5 0 5 10-5
0
5
10
v (V)
i (m
A)
Diode B is on for v > 0 and R=1kΩ.Diode A is on for v < 0 and R=2kΩ.
+
v
_
2kΩ
i
1kΩ
A B
Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.
(d)
-10 -5 0 5 10-5
0
5
10
v (V)
i (m
A)
Diode D is on for v > 0 and R=1kΩ.Diode C is on for v < 0 and R=0Ω.
+
v_
i
1kΩC
D
Problem 5: Sketch i versus v to scale for each of the circuits shown below. Assume that the diodes are ideal and allow v to range from -10 V to +10 V.
Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.
Case I: vin > 0Both diodes are on, and act as short circuits. The equivalent circuit is shown here.vo = vin
vin
+_
+
vo
_
1kΩ
1kΩ
3V
vin
+_
+
vo
_
1kΩ
Case II: vin < 0Both diodes are reverse biased and vo is the sum of the voltage drops across Zener diode and 1kΩ resistor.
vin
+_
+
vo
_
1kΩ
1kΩ
3V
vin
+_
+
vo
_
1kΩ
1kΩ
3V
Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.
Case II: vin < 0Both diodes are reverse biased and vo is the sum of the voltage drops across Zener diode and 1kΩ resistor.
vin
+_
+
vo
_
1kΩ
1kΩ
3V
vin
+_
+
vo
_
1kΩ
1kΩCase IIa: -3V < vin < 0vo = vin, because the current through Zener diode is zero, all negative voltage drop is across the Zener diode.
Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.
vin
+_
+
vo
_
1kΩ
1kΩ
Case IIb: vin < -3VExcess voltage below -3V is dropped across the two resistors (1kW and 1kW), with vo = (1/2)*(vin+3)-3= vin/2-1.5 [V].
vin
vo
1
1
2
1
-3V
-3V-3V
+_
Problem 6: Assuming ideal diodes sketch to scale the transfer characteristics (vo versus vin) for the circuit shown below.
Ia4V-
+
5V-
+(a)
Ia4V-
+
5V-
+
(a)
S
D
G
Ib
1V-
+
3V
-
+
(b)
D
S
G
Ic
4V+
-
5V
-
+
(c)
G
D
S
c
Id
3V-
+1V-
+
(d)
G
S
D