# EE111 Electric Circuit Analysis Recitation 2 September 4, 2015 1 ```EE111 Electric Circuit Analysis
Recitation 2
September 4, 2015
1. Redraw the circuit below with the minimum number of of nodes.
R4
R1
R2 R3
−
+
vs
R5
R7
R6
R8
2. For the circuit shown above:
(a) Count the number of nodes;
(b) Count the number of branches.
3. For the circuit shown below:
(a) Count the number of nodes;
(b) Count the number of branches;
(c) If we move from B to F to E to C, have we formed a path? A loop?
A
B
C
E
D
−
+
F
4. For the circuit above.
(a) If a second wire is connected between points E and D of the circuit, how many nodes
does the new circuit have?
(b) If a resistor is added to the circuit so that one terminal is connected to point C and
the other terminal is left floating, how many nodes does the new circuit have?
(c) Which of the following represents loops:
i. Moving from A to B to C to D to E to A;
ii. Moving from B to E to A;
iii. Moving from B to C to D to E to B;
iv. Moving from A to B to C to B to A.
5. Find ix in each of the circuits shown below:
EE111 Electric Circuit Analysis
Recitation 2, Page 2 of 10
September 4, 2015
4A
−
+
5V
1A
ix
(a)
ix
5A
1A
1A
(b)
1Ω
2A
5Ω
1Ω
5Ω
(c)
6. For the figure below:
(a) Find ix if iy = 2 A and iz = 0 A;
(b) Find iy if ix = 2 A and iz = 2 iy ;
(c) Find ix = iy = iz .
5A
3A
ix
iy
iz
7. Based on the figure below, find ix and iy .
ix
EE111 Electric Circuit Analysis
Recitation 2, Page 3 of 10
September 4, 2015
5A
1A
5Ω
iy
1Ω
ix
8. Based on the figure below:
(a) Calculate vy if iz = −3 A.
(b) What voltage source must replace the 5 V source to obtain vy = −6 V if iz = 0.5 A?
2Ω
+
−
+
5V
+
vx
vy
3 vx
2Ω
−
iz
1Ω
−
9. Based on the figure below:
(a) If ix = 5 A, find v1 and iy .
(b) If v1 = 3 V, find ix and iy .
(c) What value of is will lead to v1 6= v2 ?
10 Ω
10 Ω
+
is
v1
10 Ω
−
ix
iy
+
v2
10 Ω
10 Ω
−
10. Find R and G in the circuit below, if the 5 A source is supplying 100 W and the 40 V source
is supplying 500 W.
EE111 Electric Circuit Analysis
Recitation 2, Page 4 of 10
September 4, 2015
-110 V
−
+
R
G
40 V
−
+
5A
6A
11. Based on the two circuits below, find the current labeled i.
(a) Circuit one:
−
+
10 Ω
2V
−
+
1V
−
+ 3.5 V
i
10 Ω
(b) Circuit two:
2V
1Ω
1Ω
−
+
1Ω
i
2V
1Ω
−
+
6V
12. Use Ohm’s law and Kirchhoff’s laws on the circuit below to find:
(a)
(b)
(c)
(d)
vx ;
iin ;
Is ;
the power provided by the dependent source.
4Ω
+
vx
−
Is
iin
+
2V
−
+
2Ω
6A
8V
−
2Ω
4 vx
−
+
−
+ 10 V
EE111 Electric Circuit Analysis
Recitation 2, Page 5 of 10
September 4, 2015
13. (a) Use Kirchhoff’s and Ohm’s laws in a step-by-step procedure to evaluate all the currents
and voltages in the circuit shown below;
(b) Calculate the power absorbed by each element and show that the sum is zero.
−
+
v1 =60 V
5Ω
−
+
−i2 − −i3
i1
v3 =5 i2
v2
−i5 −
+
v1
4
v4
5Ω
−
+
v5
+
14. With reference to the circuit below, find the power absorbed by each of the seven circuit
elements.
1.5 Ω
2Ω
2.5 Ω
−
−20 V 14 Ω
4A
Is
4Ω
+
15. A certain circuit contains six elements and four nodes, numbered 1, 2, 3 and 4. Each circuit
element is connected between a different pair of nodes. The voltage v12 (+ reference at first
named node) is 12 V, and v34 = −8 V. Find v13 , v23 and v24 if v14 equals:
(a) 0;
(b) 6 V;
(c) -6 V.
16. Find the power being absorbed in the element X, which is a part of the circuit shown below,
if X is:
a 100 Ω resistor;
40 V independent voltage source, + reference on top;
dependent voltage source labeled 25 ix , + reference on top;
dependent voltage source labeled 0.8 v1 , + reference on top;
2 A independent current source, arrow directed upwards.
30 Ω
40 Ω
+
ix
120 V
v1
20 Ω
−
−
+
(a)
(b)
(c)
(d)
(e)
X
10 Ω
20 V
EE111 Electric Circuit Analysis
Recitation 2, Page 6 of 10
September 4, 2015
17. Find i1 in the circuit below if the dependent voltage source is labeled:
(a) 2 v2 ;
(b) 1.5 v3 ;
(c) −15 i1 .
10 Ω
−
+
−
v2
+
40 Ω
90 V
−
+
+
i1
v3
20 V
18. Refer to the circuit above, and label the dependent source 1.8 v3 . Find v3 if:
(a) the 90 V source generates 180 W;
(b) the 90 V source absorbs 180 W;
(c) the dependent source generates 100 W;
(d) the dependent source absorbs 100 W.
19. Find the power absorbed by each of the six elements shown in the circuit below.
5Ω
+
−
+
v2
20 Ω
−
+
v3
−
−
+
40 V
v1
25 Ω
−
+
−
+
4 v1 − v2
2 v3 + v2
20. Find the power absorbed by each of the six elements shown in the circuit below, if the
control for the dependent source is:
(a) 0.8 ix ;
(b) 0.8 iy .
In each case, demonstrate that the absorbed power quantities sum to zero.
EE111 Electric Circuit Analysis
Recitation 2, Page 7 of 10
5A
September 4, 2015
iy
10 mS
ix
40 mS
21. Find ix in the circuit shown below.
5 kΩ
20 kΩ
i1
3 i1
4 mA
ix
22. Find the power absorbed by each element in the single-node-pair circuit shown below. Test
your results by calculating the total sum of all power absorbed by the elements.
6Ω
7A
12 Ω
8A
4Ω
23. In the circuit shown below, assume v = 6 V, then find is .
−
v
+
5Ω
1Ω
is
2Ω
EE111 Electric Circuit Analysis
Recitation 2, Page 8 of 10
September 4, 2015
24. Using source combinations, compute i for the circuit below.
+
v
1A
1Ω
3A
3A
− i
25. Compute v for the circuit above by first combining the sources.
26. Using source combinations, compute i for the circuit below.
−
+
10 V
−
+
12 V
1 kΩ
−
+ 2V
i
27. Compute the power absorbed by each element of the circuit below and verify that their sum
is zero.
−
+
5V
−
+
10 V
16 kΩ
2A
16 Ω
7A
28. In the circuit shown below choose v1 to obtain ix = 2 A.
1Ω
1Ω
−
+
ix
-1.5 V
v1
−
+
−
+
3V
3A
1Ω
-2 A
EE111 Electric Circuit Analysis
Recitation 2, Page 9 of 10
September 4, 2015
29. For the circuit shown below:
(a) Compute the equivalent resistance.
(b) Derive an expression for the equivalent resistance if the circuit is extended using N
branches, each branch has one more resistor than the branch to its left.
1Ω
1Ω
1Ω
1Ω
1Ω
1Ω
30. Find Req for the circuit shown below.
5Ω
50 Ω
10 Ω
24 Ω
Req
60 Ω
20 Ω
40 Ω
31. Based on the network shown below:
(a) Let R = 80 Ω and find Req ;
(b) Find R if Req = 80 Ω;
(c) Find R if R = Req .
Req
10 Ω
100 Ω
R
30 Ω
40 Ω
20 Ω
EE111 Electric Circuit Analysis Recitation 2, Page 10 of 10
September 4, 2015
32. The Wheatstone bride is a well known electrical circuit which is used to measure resistance.
In the figure below, R is the unknown resistance to be measured; R1 , R2 and R3 are resistors
of known resistance and the resistance of R3 is adjustable. If the ratio of the two resistances
in the known leg (R1 /R3 ) is equal to the ratio of the two in the unknown leg (R/R2 ), then
the voltage between the two midpoints (B and D) will be zero and no current will flow
through the ammeter. If the bridge is unbalanced, the direction of the current indicates
whether R3 is too high or too low. R3 is varied until there is no current through the
R2
Using KCL and KVL, show that R = R
R3 .
1
Hints: The value of Vs is irrelevant; with im = 0, i1 = i3 and i2 = iR ; and there is no
voltage drop across the ammeter.
A
R
1
R2
−
+
vs
D
i1
i2
im
A
B
R
R3
i3
iR
C
33. Use current and voltage division on the circuit shown below to find an expression for:
(a) v2 ;
(b) v1 ;
(c) i4 .
R1
+
−
+
vs
v1
R3
−
+
v2
−
R2
R4
i4
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