Midterm Examination: Examples
Chapter 01 Basic Concepts
◼
Energy and Passive Sign Convention
1. The element shown satisfies v(i) = 2i + 1 [V]. (a) Find the net energy for -∞ < t < 8 [s] if v = 0 [V] for t < 0 [s]
and v = 1/2 [V] for t ≥ 0 [s]. (b) Is this device active or passive?
<Solution>
i
v - 1 - 1 / 2 [A] for t < 0 [ s ]
(a) i =
=
v
- 1 / 4 [A] for t  0 [ s ]
2

8
0
8
w =  vidt =  0  (-1 / 2)dt +  (1 / 2)  (-1 / 4)dt = -1 [J]
+
-
-
0
(b) Since the passive sign convention is violated and w < 0, the device is passive.
2. An element satisfies i(v) = -0.2v + 2 [A]. (a) Find the net energy for - < t < 3 [s] when i = -1[A] for t < 1 [s]
and i = 1 [A] for t ≥ 1 [s]. (b) Is this device active or passive?
<Solution>
i(t)
2 - i 15 [V] for t < 1 [ s ]
(a) v =
=
+
0.2  5 [V] for t  1 [ s ]
v
3
1
-1
-1
3
w =  vidt =  15  (-1)dt +  5 1dt = -30 + 10 = -20 [J]
-
1
(b) Since the passive sign convention is not violated and w < 0, the device is active.
Kyunghee University
Laser Photonics Laboratory
Midterm Examination: Examples
Chapter 01 Basic Concepts
◼
Proportionality and Voltage Division
If the dissipated power by 3 [] is 3 [W], find (a) R and (b) the power supplied by the voltage source.
2.5[]
10[V] +
-
2[]
1[]
R
3[]
v
+
<Solution>
(a) Voltage across 3 : p = v2/R→3 = |v32|/3→|v3| = 3 V
Voltage across 2 : Proportionality with v3, v2 = 2 V
→ Voltage across v1 + vR = v3 + v2 = 3 + 2 = 5 V
∴R=4
(b) Resistance of right side of 2.5  = ( + 4)//(2 + 3) = 2.5 
Resistance seen by voltage: Rs = 2.5 + 2.5 = 5 
∴ P = 102/Rs = 20 W
Kyunghee University
Laser Photonics Laboratory
Midterm Examination: Examples
Chapter 01 Basic Concepts
◼
Proportionality and Current Division
Find v, R, dissipated power by R, and the equivalent resistance seen by the current source.
4[]
2[A]
3[A]
5[]
R
v
+
<Solution>
Current through R: Current division, i = 1 A
(1) Voltage across R: v = -(10 - 4) = -6 V
(2) Resistance of R: R = v/i = 6/1= 6 
(3) Dissipated power by R: P = vi = 6 W or P = v2/R = 6 W or P = Ri2 = 6 W
(4) The equivalent resistance by the current source: Req = 5//4+6) = 10/3 
Kyunghee University
Laser Photonics Laboratory
Midterm Examination: Examples
Chapter 01 Basic Concepts
◼
Kirchhoff’s Law
Find v.
v1
+ 1V
-
1
2v
v
-
1
2v


+ 1V
-
1A
1
1
+
v
-
1A
+


<Solution>
By KCL at the left node: (v1 – v)/1 + (v1 – 1)/1 = 2v → 2v1 - v = 1 (1)
By KCL at the right node: (v – v1)/1 + v/1 = 1 → -v1 + 2v = 1 (2)
From, (1) and (2): v = 1V
Kyunghee University
Laser Photonics Laboratory
Midterm Examination: Examples
Chapter 01 Basic Concepts
◼
Kirchhoff’s Law
Find v.
1V
1V
+
-
+
-
v
i3
1
1
1
1A
1
1
1
i1
v
1
1A
i2
1
<Solution>
KCL: i2 = i1 + 1 (1)
KVL: For mesh 3, 1 = 1∙(i3 - i2) + 1∙(i3 - i1) (2)
From (1) and (2), i3 - i1 = 2 (3)
For super mesh formed with meshes 1 & 2: 1∙i1 + 1∙(i1 - i3) + 1∙(i2 - i3) + 1∙i2 = 0 (4)
From (1) and (4), 2i1 - i3 = -1 (5)
From (3) & (5): i1 = 1, i2 = 2, i3 = 3 (6)
From (6) and circuit: v = 1∙(i2 - i3) + 1∙i2 = 2i2 - i3 = 1 V
Kyunghee University
Laser Photonics Laboratory
Midterm Examination: Examples
Chapter 01 Basic Concepts
Kirchhoff’s Law
Find the gain vo/vs of the transistor amplifier circuit.
2k
2k
2k
io
2k
10-3vo
+
vs +
-
+
50io vo 10k
-
v
2k
io
vs +
-
2k
10-3vo
+
◼
50io
+
vo 10k
-
<Solution>
By KCL at node v: (v – vs)/2k + v/2k + io = 0 (1), io = (v – 10-3vo)/2k (2)
From (1) and (2): 3v – vs – 10-3vo = 0 (3)
@ the output circuit: vo = -(50io x10k) (4)
From (2) and (4): v = -3x10-3vo (5)
From (3) and (5): vo = -100vs → ∴ Gain = vo/vs = -100
Kyunghee University
Laser Photonics Laboratory
Midterm Examination: Examples
Chapter 01 Basic Concepts
Thevenin and Norton Equivalent Circuits
Find the Thevenin and Norton equivalent circuits.
+
+
2
10
10
2 v
1A
+ 2V
-
1A
1A
2
2 v
1A
-
1A
-
◼
<Solution>
Short circuit current: isc = 1 + 1 + 1 = 3A
Open circuit voltage: voc = 3 x 2//2 = 3V
Equivalent resistance: RT = voc/isc = 1. Or by all source killing, RT = 2//2 = 1
+
v
+
1
+
v
- 3V
1
3A
-
Thevenin
Kyunghee University
Norton
Laser Photonics Laboratory
Midterm Examination: Examples
Chapter 01 Basic Concepts
◼
Thevenin and Norton Equivalent Circuits
Find the Thevenin equivalent circuit.
i
+
+
10V -
1
+
5i
v
-
+
10V -
1
5i
i
i2 +
1
+
1
i1
v
-
<Solution>
1. Open circuit voltage:
Voltage division: i = 0 → voc = 10⸱(1/2) = 5V
Short circuit current:
KVL: 10 = 1⸱i + 5⸱i → isc = 5/3 A
Equivalent resistance: RT = voc/isc = 3
2. By KCL, KVL, and ohms law: i = -(i1 + i2) (1)
i1 + 10 - 5i = v → i1 = v + 5i -10 (1)
i2⸱1 = v → i2 = v (3)
→ v = -3i + 5 [V]
Kyunghee University
3
5V +
-
i
+
v
-
Laser Photonics Laboratory