Team Number:
Circuit Lab C Test
Kentucky Science Olympiad 2019
at the
Union College Barbourville, KY
Prepared By: Dr. Sunil Karna
School Name:
School Number:
Team Number:
Student Name:
Place:
Score:
Time: 50 minutes
BOX ALL ANSWERS
Please show your work and write your answers in the
answer sheet only.
This test is long and has different layers of difficulties.
Tiebreakers will be marked with (T1) for the first
tiebreaker, (T2) for the second and so on.
The point value for each question is marked in front of
questions.
The test will be broken up into 2 sections, a lab section,
and a general questions section.
Good luck!
Team Number:
Q.1) What is an SI unit of conductance?
[2points]
Q.2) What is a SHUNT in an electrical circuit?
[2points]
Q.3) What is the value of a color coded resistor if the color code is red, green,
orange, and silver?
[3points]
Q.4) When you pay your electric bill, what exactly are you paying for?
Page 2
[3points]
Team Number:
Q.5) Find the current through each of the resistors in the figure below.
[3points]
10V
R1 = 100Ω
R3 = 300Ω
R2 = 200Ω
Q.6) What is the equivalent resistance of the circuit below.
1Ω
3Ω
6.4Ω
2Ω
10
Req
Ω
A
5Ω
B
1Ω
5Ω
Page 3
[5points]
Team Number:
Q.7) Convert a Y -mesh into a ∇ - mesh of a blackbox. Suppose a ∇ mesh [figure 1(b)] is constructed from resistors RA = 20Ω, RB = 30Ω, and
RC = 50Ω of a Y -mesh [figure 1(a)]. Find the resistances of each arms of a
∇ - mesh.
[5points]
RAB
B
A
RAC
R
A
A
C
RB
RAC
C
=⇒
(a)
RB
RC
C
R AB
RBC
B
(b)
Figure 1: Network Conversion
Page 4
Team Number:
Q.8) Determine the equivalent resistance Req across the terminals(A,B). Assume that the resistance of each resistor is R.
[5points]
A
B
A. 65 R
B. 13 R
C. 65 R
D. 3R
Page 5
E. 6R
Team Number:
Q.9) Match each of the capacitor networks on the left with its equivalent
capacitance on the right.
[10points]
C
i.
a.
C
3
C
ii.
C
C
b.
3C
iii.
C
2C
c.
C
2
d.
2C
3
e.
2C
C
C
iv.
C
C
C
v.
C
Page 6
Team Number:
Q.10) Determine the current Ix in the following circuit.
1kΩ
4kΩ
Ix
9V
A. 23 mA
B. 31 mA
+
−
2kΩ
C. 3mA
D. 6mA
Page 7
+
−
6V
E. 5mA
[5points]
Team Number:
Q.11) Use Voltage Division to determine the voltage Vx in the following
circuit.
[5points]
2kΩ
18V
+
−
9kΩ
+
vx
6kΩ
A. 9.0V
B. 6.0V
C. 3.0V
D. 4.5V
Page 8
−
3kΩ
E. 5.0V
Team Number:
Q.12) Determine the value of the current Ix through the voltage source in
the given circuit.
[5points]
10Ω
Ix
60V
A. 5.0A
B. 10.0A
+
−
0.5Ix
15Ω
50Ω
C. 2.4A
40Ω
D. 1.6A
Page 9
E. 1.0A.
Team Number:
Q.13) Use source transformations to simplify the following circuit.
2mA
3kΩ
A
2kΩ
6V
+
−
B
A
A.
3.25mA
A
B.
1.5kΩ
2mA
6
5 kΩ
B
B
A
C.
0.4mA
A
D.
5kΩ
0.4mA
B
B
A
E.
0.8mA
5kΩ
5kΩ
B
Page 10
[5points]
Team Number:
Q.14) (T2 ) A galvanometer is being converted into a multimeter as shown in
figures below. The galvanometer has an internal resistance of 10Ω and reads
full scale when 1mA is flowing through it.
+
g
A
−
g
g
Ia
−
+
G
(a)
V
−
+
Ia
s
(b)
R
(c)
i. What is the minimum voltage that will cause the galvanometer to read
full scale?
[2points]
ii. What shunt resistance must be used in parallel with the galvanometer
to make an ammeter that reads up to 0.1A?
[3points]
iii. What resistance must be placed in series with the galvanometer to make
a voltmeter with a maximum range of 20V ?
[3points]
Page 11
Team Number:
Q.15) (T4 ) Charging and discharging of a capactor.
i. A capacitor begins charging when switch s1 is closed at time t = 0s.
What is the voltage across the capacitor at t = 0.2s?
[2points]
s1
s2
100µF
10V
1kΩ
25Ω
50Ω
100Ω
20V
25Ω
ii. Redraw the circuit at t = 5000s when switch s1 is opened and switch s2
is closed and find the Thevenin’s equivqlent.
[5points]
Vthev =
Rthev =
iii. What is the new time constant for the crcuit?
iv. What is the voltage across the capacitor at t = ∞?
Page 12
[2points]
[2points]
Team Number:
Q.16) (T3 ) What is the voltage at Vab ?
[5points]
Vab
a
2Ω
4V
1Ω
1Ω
2V
2V
b
Page 13
2Ω
4V
Team Number:
Q.17) (T1 ) Transistor-Transistor logic gate:
i. Identify the equivalent logic gate of the given analog circuits. [hint: draw
truth table]
[2points]
Vcc
R
Y
A
B
A. AND
B. OR
C. NOR
D. NAND
E. XOR.
ii. Identify the equivalent logic gate of the given analog circuits. [hint: draw
truth table]
[2points]
Vcc
A
B
Y
R
A. AND
B. OR
C. NOR
D. NAND
Page 14
E. XOR.
Team Number:
Q.18) Logic Gates:
i. Write the Boolean expression for output Y of the circuit given below.
[2points]
A
B
Y
ii. Write the above Boolean expression of Y into the simplest form. [3points]
Page 15
Team Number:
Q.19) Create an AND gate using NOR gates.
[5points]
Q.20) A P - N junction diode D is connected in series with a resistor R and
a battery V of small voltage as shown in figure below.
+
V
−
D
R
i. What bias condition is the diode in?
[2points]
ii. What is the current going through R?
[2points]
Page 16
Team Number:
Circuit Lab C Test
ANSWER SHEET
School Name:
School #:
Team #:
Student Name:
BOX ALL ANSWERS
Please show your work and write your answers in the
answer sheet only.
This test is long and has different layers of difficulties.
Tiebreakers will be marked with (T1) for the first
tiebreaker, (T2) for the second and so on.
The point value for each question is marked in front of
questions.
The test will be broken up into 2 sections, a lab section,
and a general questions section.
Good luck!
Team Number:
Q.1)
Ans: Mho (or Siemen)
Q.2)
Ans: Small resistance connected in parallel with main circuit to bypass the current in the circuit.
Q.3)
Ans: Color codes for red, green, and orange are 2,5, and
3. The code of silver color is 10%, which is a tolerance.
Hence the value of resistance is 25 × 103 ± 0.1 Ω.
Q.4)
Ans: How much electrical energy in kw-h (killowatt-hour)
unit has been used in a previous month.
Q.5)
Ans :
N ow,
Req = 100 + 30 = 400Ω;
V
10
I=
=
= 0.025A = IR1 = IR3 ;
Req 400
But,
IR2 = 0.
Page 2
Team Number:
1Ω
3Ω
1Ω
A
A
A
Req
6.4Ω
2Ω
B
5Ω
Req
6.4Ω
1.6Ω
Req
3.28Ω
B
1Ω
1Ω
(a)
B
(b)
(c)
Figure 1:
Q.6)
Ans :
100
10 k (5 + 5) =
= 5Ω;
20
16
2 k (3 + 5) =
= 1.6Ω;
10
[f igure1(a)]
[f igure1(b)]
6.4 × 1.6
= 3.28Ω
T heref ore Req = 1+(6.4 k (1.6))+1 = 2+
6.4 + 1.6
Page 3
Team Number:
Q.7)
Ans :
RAC =
RARC + RB RC + RARB 20 × 50 + 30 × 50 +
=
RB
30
RAB =
RARC + RB RC + RARB 20 × 50 + 30 × 50 + 20 × 30
=
RC
50
RBC =
RARC + RB RC + RARB 20 × 50 + 30 × 50 + 20 × 30
=
RA
20
Q.8)
Ans: Let 6i is a current flowing through terminal A to
B, then VAB = 6iReq = 2iR + iR + 2iR = 5iR.
5
T heref ore Req = R Ω
6
A
2i
2i
i
2i
B
R
R
i
2i
i
Q.9)
Ans: i.→ c; ii.→ e;
R
iii.→ b;
Page 4
iv.→ a;
v.→ d.
Team Number:
Q.10)
Ans :
or,
In loop 1,
3I1 + 2I2 = 9 − − − − − − − − − −(i)
In loop 2,
or,
1I1 + 2I1 + 2I2 = 9
4I2 + 2I2 + 2I1 = 6
I1 + 3I2 = 3 − − − − − − − − − −(ii)
On solving eqns. (i) and (ii), we get I1 = 3mA = Ix.
I2 = 0, and
Q.11)
6
3


=
× 18 −
× 18 = 9.0 volt
6+2
3+9

Ans :
Vx = V6k −V3k
Page 5

Team Number:
Q.12)
Ans : D,
V = 0.5Ix × 10 = 5Ix
5Ix
10Ω
+
−
Ix
15Ω
+
−
60V
I=
Ix
60V
50Ω
40Ω
V
5Ix
=
= 0.1Ix
R
50
15Ω
+
−
50Ω
50Ω
0.1Ix
⇒ 60 + 2.5Ix = (15 + 25)Ix ⇒ Ix =
Ix
60V
15Ω
25Ω
−
+
+
−
Page 6
2.5Ix
60
= 1.6A
37.5
Team Number:
Q.13)
4V
+
−
3kΩ
6V
2kΩ
5kΩ
A
+
−
A
2V
+
−
0.4mA
B
(a)
A
5kΩ
B
(b)
B
(c)
Figure 2:
Ans:
C.
Q.14) i.
Ans :
V = Iag = 1mA × 10Ω = 10mV
ii.
Ans :
s
0.10
s
−3
Ia ⇒ 1×10
0.1 ⇒ s =
= 0.1
Ig =
s+g
s + 10
0.99
Ans :
V = Ig (g+R) ⇒ 20 = 1×10−3(10+R) ⇒ 10+R = 2
iii.
⇒ R = 20000 − 10 = 19990 = 19.99 kΩ
Page 7
Team Number:
Q.15) (a)
s1
100µF
10V
1kΩ
−0.2 
−t 





) = 10 1 − exp(
) =
VC (t) = Vo 1 − exp(
RC
0.1


Ans :

τ = RC = 1000 × 100 × 10−6 = 0.1 sec.
(b)
capacitor is fully charged.
s2
s2
25Ω
100µF
25Ω
50Ω
50Ω
Vab
25Ω
100Ω
25Ω
⇒
100Ω
20 = I(25 + 50 + 25) = 100I ⇒ I = 0.2A
Vab = I × 50 = 0.2 × 50 = 10V = Vth
(c)
τnew = RthC = 125 × 100 × 10−6 = 0.0125sec.
Page 8

Team Number:
Rth
Vth
+
−
100µF
(d)
VC (∞) = Vo(1 − e(−∞) = 10 volt.
Rth = 100 + (50 k 50) = 125 Ω.
Q.16)
Ans :
I1 + I2 + I3 + I4
Vab = IReq =
=
1/Req
Q.17) i.
Ans: D.
A
0
1
0
1
B
0
0
1
1
ii.
Ans: B.
Page 9
Y
1
1
1
0
4
2
1
2
+ 21 + 12 + 42
1
1
1 = 2.67
+1+1+2
Team Number:
A
0
1
0
1
B
0
0
1
1
Y
0
1
1
1
Q.18) i.
Ans :
Y = (AB)(A + B)
ii.
Ans :
Y = (AB)(A+B) = (Ā+B̄)(A+B) = ĀA+ĀB+B̄A
= 0+ ĀB + B̄A+0 = ĀB + B̄A = XOR gate
A
0
1
0
1
B
0
0
1
1
Page 10
Y
0
1
1
0
Team Number:
Q.19)
Ans :
¯ B̄
¯ = AB
Y = Ā + B̄ = Ā
A
Y
B
Q.20) i.
Ans: Reverse Bias.
ii.
Ans: zero.
Page 11