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