ES 250 First Midterm Practice Exam 2
1.
i a = _−0.333__A, i b = ___1____A,
i 2 = __0.333__A,
and
v1 = _−1.333__V
Use equivalent resistances to reduce the circuit to
From KCL i b = 4 i a + i a
⇒ i b = −3 i a .
From KVL
12 i a + 10 i b − 6 = 0 ⇒ 12 i a + 10 ( −3 i a ) = 6
So i a = −
1
A , v a = −4 A and i b = 1 A .
3
Returning our attention to the original circuit, notice that i a and i b were not changed when the circuit was
reduced. Now v1 = ( 5 || 20 ) i a = ( 4 )( −0.333) = −1.333 V and i 2 =
12
i = 0.333 A .
12 + 24 b
2.
The current in the 20-Ω resistor is i = ____−1.25___A.
a
The voltage across the 10-Ω resistor is v = ____−30_____V.
b
The (independent) voltage source current is i = __−4.25____A.
c
25
= −1.25 A .
20
From KVL 4 i a − v b + 20 i a = 0 ⇒ v b = 24 i a = 24 ( −1.25 ) = −30 V
From Ohm’s law i a = −
From KCL i c = i a +
vb
10
= −1.25 +
−30
= −4.25 A
10
The Ohmmeter measures equivalent resistance.
3.
a. To cause R eq = 12 Ω, choose R = ___16__Ω.
b. If R = 14 Ω then R eq = ___11.5____Ω.
R eq = 2 + ( 20 || ( 4 + R ) ) . When R eq = 12 Ω then
12 = 2 + ( 20 || ( 4 + R ) ) ⇒ 10 =
20 ( 4 + R )
⇒ 24 + R = 2 ( 4 + R ) ⇒ R = 16 Ω
20 + ( 4 + R )
When R = 14 Ω then
R eq = 2 + ( 20 || ( 4 + 14 ) ) = 2 + ( 20 ||18 ) = 2 +
4.
20 (18 )
= 11.4736 Ω
20 + 18
Consider this combination of resistors. Let R p denote the equivalent resistance.
(a) Suppose 40 Ω ≤ R ≤ 400 Ω. Determine the corresponding range of values of R p :
___53.33__ Ω ≤ R p ≤ __117.33___ Ω
(b) Suppose instead R = 0 (a short circuit). Then R p = ___32____ Ω
(c) Suppose instead R = ∞ (an open circuit). Then R p = ___160___ Ω
(d) Suppose instead the equivalent resistance is R p = 80 Ω. Then R = ___120___ Ω
160 ( 80 ) 160
=
= 53.33 Ω .
160 + 80
3
160 ( 440 ) 16 ( 44 )
When R = 400 Ω then R p = 160 || ( 40 + 400 ) = 160 || 440 =
=
= 117.33 Ω .
160 + 440
6
160 ( 40 ) 16 ( 4 )
When R = 0 then R p = 160 || ( 0 + 40 ) = 160 || 40 =
=
= 32 Ω .
160 + 40
2
When R = 40 Ω then R p = 160 || ( 40 + 40 ) = 160 || 80 =
When R = ∞ then R p = 160 || ( ∞ + 40 ) = 160 || ∞ =
When R p = 80 Ω then
80 = 160 || ( R + 40 ) =
1
1
1
+
160 ∞
= 160 Ω .
160 ( R + 40 )
160
⇒ R + 200 =
( R + 40 ) = 2 R + 80 ⇒ R = 120 Ω .
160 + R + 40
80
5.
Here’s a single circuit drawn in four parts for convenience. The four parts are connected by the dependent
sources. Given that i1 = 0.8 A, determine the values of R1 , v 2 , v 3 , and i 4 .
R1 = ____5____Ω , v 2 = ____−4____V, v 3 = ___2___V and i 4 = __−0.96____A.
1 ⎛ 24 ⎞
12
0.8 = ⎜
⇒ R1 + 10 =
= 15 ⇒ R1 = 5 Ω
⎟
2 ⎜⎝ R1 + 10 ⎟⎠
0.8
1
20 || 80
16
v 2 = − ⎡⎣15 ( 0.8 ) ⎤⎦ = −4 V , v 3 =
10 ( 0.8 ) =
8=2 V
3
48 + ( 20 || 80 )
48 + 16
i4 = −
6.
40 || 40
1
2.4 ( 0.8 ) = − (1.92 ) = −0.96 A
2
( 40 || 40 ) + 20
Encircled numbers are node numbers. The
corresponding node voltages are:
v1 = 12 V, v 2 = 10.5 V and v 3 = 6 V
The value of the gain of the CCCS is k = ____5.00_____ A/A.
The resistance of the resistor at the top of the circuit is R = __600_____Ω. (Round to an integer.)
The power supplied by the independent (0.1 A) current source is ____−0.6___W.
ia =
12 − 10.5
10.5 − 6
= 0.015 A , i b =
= 0.09 A
100
50
i c + i b = 0.1 ⇒ i c = −0.09 + 0.1 = 0.01 A
ia + k ia = ib
⇒ k i a = i b − i a = 0.09 − 0.015 = 0.075 A
k=
k ia
ia
=
0.075
12 − 6
6
= 5 A/A , R =
=
= 600 Ω and the power is − ( 6 )( 0.1) = −0.6 W
0.015
ic
0.01
Let i1, i2 and i3 denote the mesh currents in
meshes 1, 2 and 3, respectively.
7.
Determine the values of these mesh currents:
i1 = ____2____A and i2 = ____−3_____A
Determine the value of the resistance R:
R = _____5_______ Ω
i 2 = −3 A , i1 − i 2 = 5 ⇒ i1 = 2 A , i 3 = −1 A . The applying KVL to mesh 3 gives
2 ( i 3 − i1 ) + 4 i 3 + R ( i 3 − i 2 ) = 0 ⇒ R = −
2 ( i 3 − i1 ) + 4 i 3
i3 − i2
=−
2 ( −1 − 2 ) + 4 ( −1)
=5Ω
−1 − ( −3)