**Network Theory and Complex Circuit Diagram MCQs ( Network Theory ) MCQs – ****Network Theory ****MCQs**

**Latest Network Theory MCQs**

By practicing these MCQs of ** Network Theory and Complex Circuit Diagram MCQs ( Network Theory ) MCQs – Latest Competitive MCQs **, an individual for exams performs better than before. This post comprising of objective questions and answers related to **“** Network Theory and Complex Circuit Diagram MCQs ( Network Theory ) Mcqs **“.** As wise people believe “Perfect Practice make a Man Perfect”. It is therefore practice these mcqs of Network Theory to approach the success. Tab this page to check **” Network Theory and Complex Circuit Diagram MCQs ( Network Theory )”** for the preparation of **competitive mcqs, FPSC mcqs, PPSC mcqs, SPSC mcqs, KPPSC mcqs, AJKPSC mcqs, BPSC mcqs, NTS mcqs, PTS mcqs, OTS mcqs, Atomic Energy mcqs, Pak Army mcqs, Pak Navy mcqs, CTS mcqs, ETEA mcqs** and others.

**Network Theory MCQs – Network Theory and Complex Circuit Diagram MCQs ( Network Theory ) MCQs**

The most occurred **mcqs of Network Theory and Complex Circuit Diagram MCQs ( Network Theory )** in past papers. **Past papers of Network Theory and Complex Circuit Diagram MCQs ( Network Theory )** Mcqs. **Past papers of** ** Network Theory and Complex Circuit Diagram MCQs ( Network Theory )** Mcqs . Mcqs are the necessary part of any competitive / job related exams. The Mcqs having specific numbers in any written test. It is therefore everyone have to learn / remember the related ** Network Theory and Complex Circuit Diagram MCQs ( Network Theory )** Mcqs. The Important series of ** Network Theory and Complex Circuit Diagram MCQs ( Network Theory ) Mcqs** are given below:

# Advanced Problems on Network Theory – 1

**1. Branch current and loop current relation is expressed in matrix form shown below, where Ij represents branch current and Ik represents loop current.****[I _{1}; I_{2}; I_{3}; I_{4}; I_{5}; I_{6}; I_{7}; I_{8}] = [0 0 1 0; -1 -1 -1 0; 0 1 0 0; 1 0 0 0; 0 0 -1 -1; 1 1 0 -1; 1 0 0 0; 0 0 0 1] [I_{1}; I_{2}; I_{3}; I_{4}] The rank of the incidence matrix is?**

**a) 4**

b) 5

c) 6

d) 8

**Answer: a**

**Explanation: Number of branches b = 8**

**Number of links l = 4**

**Number of twigs t = b – l = 4**

**Rank of matrix = n – 1 = t = 4.**

**2. A capacitor, used for power factor correction in a single phase circuit decreases which of the following?**

a) Power factor

**b) Line current**

c) Both Line current and Power factor

d) Neither Line current nor Power factor

**Answer: b**

**Explanation: We know that a capacitor is used to increase the Power factor. However, with decrease in line current the power factor is increased. Hence line current decreases.**

**3. D is the distance between the plates of a parallel plate capacitor. The dielectric constants are ∈**

_{1}and ∈_{2}respectively. The total capacitance is proportional to ____________**a) ∈1∈2∈1+∈2**

b) ∈

_{1}– ∈

_{2}

c) ∈1∈2

d) ∈

_{1}∈

_{2}

**Answer: a**

**Explanation: The combination is equal to two capacitors in series.**

**So, C = [∈0∈1(A0.5d)][∈0∈2(A0.5d)]∈0∈1A0.5d+∈0∈1A0.5d**

**Hence, C is proportional to ∈1∈2∈1+∈2.**

**4. A two branch circuit has a coil of resistance R**

_{1}, inductance L_{1}in one branch and capacitance C_{2}in the second branch. If R is increased, the dynamic resistance is going to ___________a) Increase

**b) Decrease**

c) Remains constant

d) May increase or decrease

**Answer: b**

**Explanation: We know that,**

**Dynamic resistance = L1R1C2**

**So, if R**

_{1}is increased, keeping Inductance and Capacitance same, so The Dynamic resistance will decrease, as the denomination is increasing.**5. A 1 μF capacitor is connected to 12 V batteries. The energy stored in the capacitor is _____________**

a) 12 x 10

^{-6}J

b) 24 x 10

^{-6}J

c) 60 x 10

^{-6}J

**d) 72 x 10**

^{-6}J**Answer: d**

**Explanation: We know that,**

**Energy, E = 0.5 CV**

^{2}**= 0.5 X 1 X 10**

^{-6}X 144**= 72 x 10**

^{-6}J.**6. For the two circuits shown below, the relation between I**

_{A}and I_{B}is ________a) I

_{B}= I

_{A}+ 6

b) I

_{B}= I

_{A}+ 2

**c) I**

_{B}= 1.5I_{A}d) I

_{B}= I

_{A}

**Answer: c**

**Explanation: In the circuit of figure (I**

_{B}), transforming 3A source into 18 V source, all sources are 1.5 times of that in circuit (I_{A}). Hence, I_{B}= 1.5I_{A}.**7. For the circuit given below, the current I in the circuit is ________**

**a) –j1 A**

b) J1 A

c) Zero

d) 20 A

**Answer: a**

**Explanation: X**

_{EQ}= sL + R×1/sCR+1/sC=sL+R1+sRC**I**

_{O}= VXEQ**∴ I = XCXC+R I**

_{O}**= 1/sC1sC+R×VsL(1+sRC)+R(1+sRC)**

**= 11+sRC×VsL(1+sRC)+R(1+sRC)**

**= VsL(1+sRC)+R**

**= Vj×103×20×10−3(1+j×103×50×10−6+1)**

**= V20j(1+j50×10−3)+1**

**= V20j−1+1=2020j = -j1 A.**

**8. An AC source of RMS voltage 20 V with internal impedance Z**

_{S}= (1+2j) Ω feeds a load of impedance Z_{L}= (7+4j) Ω in the circuit given below. The reactive power is _________a) 8 VAR

**b) 16 VAR**

c) 28 VAR

d) 32 VAR

**Answer: b**

**Explanation: Current I = VZL+ZS=20∠0°8+6j**

**= 2082+62√=∠0°∠arctan(34)**

**= 2010 ∠-arc tan(34)**

**= 2∠-arc tan(34)**

**Power consumed by load = |I|**

^{2}Z_{L}**= 4(7+4j)**

**= 28 + j16**

**∴ The reactive power = 16 VAR.**

**9. In the circuit given below, R**

_{I}= 1 MΩ, R_{O}= 10 Ω, A = 10^{6}and V_{I}= 1μV. Then the output voltage, input impedance and output impedance respectively are _________**a) 1 V, ∞ and 10 Ω**

b) 1 V, 0 and 10 Ω

c) 1V, 0 and ∞

d) 10 V, ∞ and 10 Ω

**Answer: a**

**Explanation: V**

_{O}(output voltage) = AV_{I}= 10^{6}× 10^{-6}= 1 V**V**

_{1}= Z_{11}I_{1}+ Z_{12}I_{2}**V**

_{2}= Z_{21}I_{1}+ Z_{22}I_{2}**Here, I**

_{1}= 0**Z**

_{11}= V1I1=VO0 = ∞**Z**

_{22}= V2I2=AVII2**Or, Z**

_{22}= 1I2 = R_{O}= 10 Ω.**10. If operator ‘a’ = 1 ∠120°. Then (1 – a) is equal to ____________**

a) 3–√

**b) 3–√∠-30°**

c) 3–√∠30°

d) 3–√∠60°

**Answer: b**

**Explanation: Given that, ‘a’ = 1 ∠120°**

**So, 1 – a = 1 – 1∠120°**

**= 1 + 0.5 – j 0.866**

**= 1.5 – j 0.866**

**= 3∠-30°.**

**11. For making a capacitor, the dielectric should have __________**

**a) High relative permittivity**

b) Low relative permittivity

c) Relative permittivity = 1

d) Relative permittivity neither too high nor too low

**Answer: a**

**Explanation: Relative permittivity is for ideal dielectric which is air. Achieving such a precise dielectric is very difficult.**

**Low relative permittivity will lead to low value of capacitance.**

**High relative permittivity will lead to a higher value of capacitance.**

**12. In the circuit shown below, the voltage V will be __________**

**a) – 3V**

b) Zero

c) 3 V

d) 5 V

**Answer: a**

**Explanation: By applying KVL, I = 1 A**

**V**

_{AB}– 2 × 1 + 5 = 0**Or, V**

_{AB}= -3 V.**13. If A = 3 + j1, then A**

^{4}is equal to __________a) 3.16 ∠18.4°

**b) 100 ∠73.72°**

c) 100 ∠18.4°

d) 3.16 ∠73.22°

**Answer: b**

**Explanation: Given A = 3 + j1**

**So, 3 + j1 = 10∠18.43°**

**Or, 3 + j1 = (10)**

^{4}∠4 X 18.43°**= 100∠73.72°.**

**14. In the figures given below, Value of R**

_{A}, R_{B}and R_{C}are 20 Ω, 10 Ω and 10 Ω respectively. The resistances R_{1}, R_{2}and R_{3}in Ω are ________**a) 2.5, 5 and 5**

b) 5, 2.5 and 5

c) 5, 5 and 2.5

d) 2.5, 5 and 2.5

**Answer: a**

**Explanation: R**

_{1}= RBRCRA+RB+RC=10040 = 2.5 Ω**R**

_{2}= RARCRA+RB+RC=20040 = 5 Ω**R**

_{3}= RBRARA+RB+RC=20040 = 5 Ω.**15. The resistance of a thermistor decreases with increases in __________**

a) temperature

**b) circuit**

c) light control

d) sensors

**Answer: b**

**Explanation: The resistance of a thermistor decreases with increases in temperature. Hence, it is used to monitor hot spot temperature of electric machines.**

# Advanced Problems on Network Theory – 2

**1. In the circuit given below, the value of R is _________**

a) 10 Ω

b) 18 Ω

c) 24 Ω**d) 12 Ω****Answer: d****Explanation: By KCL,****∴ VP−401+VP−10014+VP2 = 0****Or, 22 V _{P} = 660**

**∴ V**

_{P}= 30 V**Potential difference between node x and y = 60 V**

**∴ -I – 5 + 40−30I = 0**

**Or, I = 5 A**

**∴ R = 605= 12 Ω.**

**2. In the circuit given below, the resistance between terminals A and B is 7Ω, between terminals B and C is 12Ω and between terminals C and A is 10Ω. The remaining one terminal in each case is assumed to be open. Then the value of R**

_{A}and R_{B}are _________a) R

_{A}= 9 Ω and R

_{B}= 7 Ω

**b) R**

_{A}= 2.5 Ω and R_{B}= 4.5 Ωc) R

_{A}= 3 Ω and R

_{B}= 3 Ω

d) R

_{A}= 5 Ω and R

_{B}= 1 Ω

**Answer: b**

**Explanation: Given R**

_{A}+ R_{B}= 7 with C open**R**

_{B}+ R_{C}= 12 with A as open**R**

_{A}+ R_{C}= 10 with B open**Then, R**

_{A}+ R_{B}+ R_{C}= 292 = 14.5**Hence, R**

_{A}= 2.5 Ω, R_{B}= 4.5 Ω and R_{C}= 7.5 Ω.**3. Currents I**

_{1}, I_{2}and I_{3}meet at a junction in a circuit. All currents are marked as entering the node. If I_{1}= -6sin(ωt) mA and I_{2}= 8 cos(ωt) mA, the I_{3}is ________a) 10 cos(ωt + 36.87) mA

b) 14 cos(ωt + 36.87) mA

c) -14 sin(ωt + 36.87) mA

**d) -10 cos(ωt + 36.87) mA**

**Answer: d**

**Explanation: Applying KCL, we get, I**

_{1}+ I_{2}+ I_{3}= 0**∴ -6 sin(ωt) + 8 cos(ωt) + I**

_{3}= 0**∴ I**

_{3}= 6 sin (ωt) – 8 cos (ωt)**= 10[sin (ωt).sin (36.86) – cos (ωt) cos (36.86)]**

**=-10[cos (ωt) cos (36.86) – sin (ωt) sin (36.86)]**

**= -10 cos (ωt + 36.86)**

**[As, cos (A+B) = cosA.cosB – sinA.sinB].**

**4. Viewed from the terminals A and B the circuit given below can be reduced to an equivalent circuit with a single voltage source in series with a resistor with ________**

a) 5 V source in series with 10 Ω resistor

**b) 1 V source in series with 2.4 Ω resistor**

c) 15 V source in series with 2.4 Ω resistor

d) 1 V source in series with 10 Ω resistor

**Answer: b**

**Explanation: Applying Thevenin’s Theorem R**

_{EQ}= 6 || 4**= 6×46+4**

**= 2410 = 2.4 Ω**

**V**

_{AB}= 10 – 6 × (1510) = 1 V.**5. In the circuit given below, the voltage across the 2Ω resistor is ________**

a) 3.41 V

**b) -3.41 V**

c) 3.8 V

d) -3.8 V

**Answer: b**

**Explanation: Applying KCL to node A, VA−1010+VA20+VA7 = 0**

**Or, V**

_{A}(0.1 + 0.05 + 0.143) = 1**Or, V**

_{A}= 3.41 V**The voltage across the 2 Ω resistor due to 10 V source is V**

_{2}= VA7×2 = 0.97 V**V**

_{2Ω}due to 20 V source, VA10+VA20+VA−207 = 0**Or, 0.1 V**

_{A}+ 0.05V_{A}+ 0.143V_{A}= 2.86**∴ V**

_{A}= 2.860.293 = 9.76 V**V**

_{2Ω}= VA−207 × 2 = -2.92 V**The current in 2 Ω resistor = 2 × 55+8.67**

**= 1013.67 = 0.73 A**

**The voltage across the 2 Ω resistor = 0.73 × 2 = 1.46 V**

**V**

_{2Ω}= 0.97 – 2.92 -1.46 = -3.41 V.**6. In the circuit given below, the value of I**

_{X}using nodal analysis is _______a) -2.5 A

**b) 2.5 A**

c) 5 A

d) -5 A

**Answer: b**

**Explanation: Applying KCL, we get, I**

_{1}+ 5 = I_{2}+ I_{3}**∴ 10–V11+5=V12+V26**

**∴ 30 – 3V**

_{1}+ 15 = 3V_{1}+ V_{2}**∴ 6 V**

_{1}+ V_{2}= 45**From voltage source, V**

_{2}– V_{1}= 10**Now, 7 V**

_{1}= 35, V_{1}= 5 V**And V**

_{2}= 15 V**∴ I**

_{X}= V26=156 = 2.5 A.**7. In the circuit given below, the values of V**

_{1}and V_{2}respectively are _________a) 0 and 5V

b) 5 and 0 V

c) 5 and 5 V

**d) 2.5 and 2.5 V**

**Answer: d**

**Explanation: I**

_{1}= V1−V22**Applying KCL at node 1, 5 = V11+V1−V22+V1**

**10 = 2V**

_{1}+ V_{1}– V_{2}+ 2V_{1}**Or, 10 = 5V**

_{1}– V_{2}**KCL at node 2, V1−V22 + V**

_{1}+ 2I_{1}= V_{2}**∴ 1.5 V**

_{1}– V_{2}= 0**∴ V**

_{1}= V_{2}= 2.5 V.**8. In the circuit given below, the voltage across the 18 Ω resistor is 90 V. The voltage across the combined circuit is _________**

**a) 125 V**

b) 16 V

c) 24 V

d) 40 V

**Answer: a**

**Explanation: Current through the 18 Ω resistance = 9018 = 5 A**

**Equivalent resistance of 3 Ω and 7 Ω banks = 3×63+6 = 2 Ω**

**Since, this 2 Ω resistance is in series with 18 Ω resistance, therefore total resistance = 18 + 2 = 20 Ω**

**This 20 Ω resistance is in parallel with 5 Ω resistance = 5×205+20 = 4 Ω**

**Hence, total resistance of the circuit = 1 + 4 = 5 Ω**

**Current through this branch = 5 A**

**∴ Voltage across dc= 5 × 20 = 100 V**

**Hence current through 5 Ω resistance = 1005 = 20 A**

**∴ Total current = 20 + 5 = 25 A**

**Since, total resistance of the circuit is 5 Ω therefore, voltage E = 25 × 5 = 125 V.**

**9. In the circuit given below, the value of R in the circuit, when the current is zero in the branch CD is _________**

a) 10 Ω

b) 20 Ω

c) 30 Ω

**d) 40 Ω**

**Answer: d**

**Explanation: The current in the branch CD is zero if the potential difference in the branch CD is zero.**

**That is, V**

_{C}= V_{D}**Or, V**

_{10}= V_{C}= V_{D}= V_{A}× 1015**V**

_{R}= V_{A}× R20+R**And V**

_{10}= V_{R}**∴ V**

_{A}× 1015 = V_{A}× RR+20 ∴ R = 40 Ω.

#### Elements of Realizability and Synthesis of One-Port Networks

**10. Two capacitors of 0.5 μF and 1.5 μF capacitance are connected in parallel across a 110 V dc battery. The charges across the two capacitors after getting charged is ___________**

a) 55 μC each

b) 275 μC each**c) 55 μC and 275 μC respectively**

d) 275 μC and 55 μC respectively**Answer: c****Explanation: Q _{1} = 0.5 x 10^{-6} x 110**

**= 55 μC.**

**Also, Q**

_{2}= 2.5 x 10^{-6}x 110**= 275 μC.**

**11. Consider a circuit having resistances 2 Ω and 2 Ω in series with an inductor of inductance 2 H. The circuit is excited by a voltage of 12 V. A switch S is placed across the first resistance. Battery has remained switched on for a long time. The current i(t) after switch is closed at t=0 is _____________**

a) 6

**b) 6 – 3e**

^{-t}c) 6 + 3e

^{-t}

d) 3 – 6e

^{-t}

**Answer: b**

**Explanation: From the figure, we can infer that,**

**I(t) = 1212(1–24e−t)**

**= 6 – 3e**

^{-t}.**12. A resistance R is connected to a voltage source V having internal resistance R**

_{I}. A voltmeter of resistance R_{2}is used to measure the voltage across R. The reading of the voltmeter is _____________a) VSR1R2R2R1+R1R+RR2

b) VsRR+Rs

c) VR1R2R1R2–R1R–RR2

**d) VSRR2R1R2+R1R+RR2**

**Answer: d**

**Explanation: Effective resistance of R and R**

_{m}is**R**

_{eq}= RRmR+Rm**Therefore the reading is VR1+RR+R2[RR2R+R2]**

**= VSRR2R1R2+R1R+RR2.**

**13. When a lead acid battery is being charged, the specific gravity of the electrolyte will ___________**

a) Decrease

**b) Increase**

c) Either Increase or Decrease

d) Neither Increase nor Decrease

**Answer: b**

**Explanation: We know that the specific gravity of electrolyte is highest when battery is fully charged and is lowest when discharged. So, the specific gravity of the electrolyte will increase when a lead acid battery is being charged.**

**14. A series RLC circuit has a resonant frequency of 550 Hz. The maximum voltage across C is likely to occur at a frequency of ___________**

a) 1000 Hz

b) 2000 Hz

c) 1025 Hz

**d) 500 Hz**

**Answer: d**

**Explanation: We know that, maximum voltage across capacitance occurs at a frequency slightly less than resonant frequency.**

**Here, given that resonant frequency = 550 Hz.**

**So, out of the given options 500 is the lowest nearest integer to 550.**

**Hence, 975 Hz.**

# Advanced Problems Involving Complex Circuit Diagram – 1

**1. The current wave shape is in the form of a square terminating at t = 4sec. The voltage across the element increases linearly till t = 4 sec and then becomes constant. The element is ____________**

a) Resistance

b) Inductance**c) Capacitance**

d) Semi-conductor**Answer: c****Explanation: We know that, when a current pulse is applied to a capacitor, the voltage will have a waveform which rises linearly and then becomes constant towards the end of pulse. Hence, the element is a capacitor.****2. An infinite ladder is constructed with 1 Ω and 2 Ω resistor shown below. The current I flowing through the circuit is ___________**

a) 8.18 A

b) 0 A**c) 9 A**

d) 10 A**Answer: c****Explanation: Equivalent Resistance, R _{EQ} = R = 1 + (2 || R)**

**Or, R**

_{EQ}= 2**So, I = 182 A = 9 A.**

**3. In the circuit given below, the phase angle of the current I with respect to the voltage V**

_{1}is __________a) 0°

b) +45°

c) -45°

**d) -90°**

**Answer: d**

**Explanation: Net voltage applied to the circuit is 200∠0° V**

**I**

_{1}= 200∠0°10.0**= 20∠0° = 20**

**I**

_{2}= 200∠0°10∠90°**= 20∠-90° = -j20**

**I = I**

_{1}+ I_{2}= 20(1-j) = 202–√∠45°**Voltage V**

_{1}= 100(1+j)**= 1002–√∠45°**

**∴ Required phase angle = -45° – 45° = -90°.**

**4. Consider a circuit having 3 identical Ammeters A**

_{1}, A_{2}, A_{3}parallel to one another. The 1^{st}Ammeter is in series with a resistance, the 2^{nd}Ammeter is in series with a capacitor and the circuit is excited by a voltage V. If A_{1}and A_{3}read 5 and 13 A respectively, reading of A_{2}will be?a) 8 A

b) 13 A

c) 18 A

**d) 12 A**

**Answer: d**

**Explanation: We can infer from the circuit,**

**A**

_{2}= 132–52−−−−−−√**Or, A**

_{2}= 169–25−−−−−−√**Or, A**

_{2}= 144−−−√**Or, A**

_{2}= 12 A.**5. In the circuit given below, the value of V**

_{1}is __________a) 32.2 V

**b) -25.23 V**

c) 29.25 V

d) -29.25 V

**Answer: b**

**Explanation: VA30+VA−4012+VA−VB8 = 0**

**Or, 29 V**

_{A}– 15 V_{B}= 400**Also, VB−VA8+VB−1208 + 6 = 0**

**Or, V**

_{A}= 65.23 V, V_{B}= 99.44 V**V**

_{1}= 40-65.23 = -25.23 V.**6. For the three coupled coils shown in figure, KVL equation is ____________**

a) V = (L

_{1}+ L

_{2}+ L

_{3}) didt

b) V = (L

_{1}– L

_{2}– L

_{3}– M

_{13}) didt

**c) V = (L**

_{1}+ L_{2}+ L_{3}+ 2M_{12}– 2M_{23}– 2M_{13}) didtd) V = (L

_{1}+ L

_{2}+ L

_{3}– 2M

_{12}+ 2M

_{23}+ 2M

_{13}) didt

**Answer: c**

**Explanation: M**

_{12}is positive while M_{23}and M_{13}are negative because of dots shown in figure.**So, the KVL equation is given by,**

**V = (L**

_{1}+ L_{2}+ L_{3}+ 2M_{12}– 2M_{23}– 2M_{13}) didt.**7. A circuit excited by voltage V has a resistance R which is in series with an inductor and capacitor, which are connected in parallel. The voltage across the resistor at the resonant frequency is ___________**

**a) 0**

b) V2

c) V3

d) V

**Answer: a**

**Explanation: Dynamic resistance of the tank circuit, Z**

_{DY}= LRLC**But given that R**

_{L}= 0**So, Z**

_{DY}= L0XC = ∞**Therefore current through circuit, I = V∞ = 0**

**∴ V**

_{D}= 0.**8. In the circuit given below, the value of resistance R is _________**

a) 10 Ω

b) 18 Ω

c) 24 Ω

**d) 12 Ω**

**Answer: d**

**Explanation: Using KVL in loop 1, we get, 100 – 14I – 30 = 0**

**Or, I = 7014 = 5 A**

**Then, V**

_{P}– V_{Q}= 14I – (15-I).1**= 70 – 10 = 60 V**

**∴ R = 6010−I=605 = 12 Ω.**

**9. The current flowing through the resistance R in the circuit in the figure has the form 2 cos 4t, where R is ____________**

a) (0.18 + j0.72)

b) (0.46 + j1.90)

c) – (0.18 + j1.90)

**d) (0.23 – 0.35 j)**

**Answer: d**

**Explanation: Inductor is not given, hence ignoring the inductance. Let I**

_{1}and I_{2}are currents in the loop then,**I**

_{1}= 2cos4t3**= 0.66 cos 4t**

**Again, I**

_{2}= −jX4X0.75I13.92−2.56j**= (0.23 – 0.35j) cos 4t**

**10. In the circuit given below, the voltage V**

_{AB}is _________**a) 6 V**

b) 25 V

c) 10 V

d) 40 V

**Answer: a**

**Explanation: For finding the Thevenin Equivalent circuit across A-B we remove the 5 Ω resistor.**

**Then, I = 10+5015 = 4 A**

**V**

_{OC}= 50 – (10×4) =10 V**And R**

_{EQ}= 10×510+5=103 Ω**Current I**

_{1}= 1010/3+5=65**Hence, V**

_{AB}= 65×5 = 6 V.**11. In the circuit given below, the magnitudes of V**

_{L}and V_{C}are twice that of V_{K}. Calculate the inductance of the coil, given that f = 50.50 Hz.a) 6.41 mH

b) 5.30 mH

**c) 3.18 mH**

d) 2.31 mH

**Answer: c**

**Explanation: V**

_{L}= V_{C}= 2 VR**∴ Q = VLVR = 2**

**But we know, Q = ωLR=1ωCR**

**∴ 2 = 2πf×L5**

**Or, L = 3.18 mH.**

**12. In the circuit given below, the current source is 1 A, voltage source is 5 V, R**

_{1}= R_{2}= R_{3}= 1 Ω, L_{1}= L_{2}= L_{3}= 1 H, C_{1}= C_{2}= 1 F. The current through R_{3}is _________a) 1 A

**b) 5 A**

c) 6 A

d) 8 A

**Answer: b**

**Explanation: At steady state, the circuit becomes,**

**∴ The current through R**

_{3}= 51 = 5 A.**13. In the circuit given below, the capacitor is initially having a charge of 10 C. 1 second after the switch is closed, the current in the circuit is ________**

**a) 14.7 A**

b) 18.5 A

c) 40.0 A

d) 50.0 A

**Answer: a**

**Explanation: Using KVL, 100 = Rdqdt+qC**

**100 C = RCdqdt + q**

**Or, ∫qqodq100C−q=1RC∫t0dt**

**100C – q = (100C – q**

_{o})e^{-t/RC}**I = dqdt=(100C–qo)RCe−1/1**

**∴ e**

^{-t/RC}= 40e^{-1}= 14.7 A.**14. A circuit is given in the figure below. The Norton equivalent as viewed from terminals x and x’ is ___________**

a) 6 Ω and 1.333 A

**b) 6 Ω and 0.833 A**

c) 32 Ω and 0.156 A

d) 32 Ω and 0.25 A

**Answer: b**

**Explanation: We, draw the Norton equivalent of the left side of xx’ and source transformed right side of yy’.**

**V**

_{xx’}= V_{N}= 48+82418+124 = 5V**∴ R**

_{N}= 8 || (16 + 8)**= 8×248+24 = 6 Ω**

**∴ I**

_{N}= VNRN=56 = 0.833 A.**15. In the circuit given below, the current source is 1 A, voltage source is 5 V, R**

_{1}= R_{2}= R_{3}= 1 Ω, L_{1}= L_{2}= L_{3}= 1 H, C_{1}= C_{2}= 1 F. The current through the voltage source V is _________a) 1 A

b) 3 A

c) 2 A

**d) 4 A**

**Answer: d**

**Explanation: At steady state, the circuit becomes,**

**∴ The current through the voltage source V = 5 – 1 = 4 A.**

# Advanced Problems Involving Complex Circuit Diagram – 2

**1. In the circuit given below, the KVL for first loop is ___________**

a) V (t) = R_{1}i_{1} + L_{1} di1dt + M di2dt

b) V (t) = R_{1}i_{1} – L_{1} di1dt – M di2dt**c) V (t) = R _{1}i_{1} + L_{1} di1dt – M di2dt**

d) V (t) = R

_{1}i

_{1}– L

_{1}di1dt + M di2dt

**Answer: c**

**Explanation: We know that, in general, the KVL is of the form V (t) = R**

_{1}i_{1}+ L_{1}di1dt + M di2dt**But here, M term is negative because i**

_{1}, is entering the dotted terminal and i_{2}, is leaving the dotted terminal.**So, V (t) = R**

_{1}i_{1}+ L_{1}di1dt – M di2dt.**2. In a parallel RL circuit, 12 A current enters into the resistor R and 16 A current enters into the Inductor L. The total current I the sinusoidal source is ___________**

a) 25 A

b) 4 A

**c) 20 A**

d) Cannot be determined

**Answer: c**

**Explanation: Currents in resistance and inductance are out of phase by 90°.**

**Hence, I = I21+I22**

**Or, I = [12**

^{2}+ 16^{2}]^{0.5}**Or, I = 144+256−−−−−−−−√=400−−−√**

**= 20 A.**

**3. Consider a series RLC circuit having resistance = 1Ω, capacitance = 1 F, considering that the capacitor gets charged to 10 V. At t = 0 the switch is closed so that i = e**

^{-2t}. When i = 0.37 A, the voltage across capacitor is _____________a) 1 V

**b) 6.7 V**

c) 0.37 V

d) 0.185 V

**Answer: b**

**Explanation: We know that, during discharge of capacitor,**

**V**

_{C}= V_{R}**Now, V**

_{R}= 0.67 X 10 = 6.7 V**So, V**

_{C}= 6.7 V.**4. A waveform is of the form of a trapezium, which increases linearly with the linear slope till θ = π3, constant till θ = π2 and again linearly decreases to 0 till θ = π. The average value of this waveform is ______________**

a) 2 V

b) 0 V

c) 4 V

**d) 3 V**

**Answer: d**

**Explanation: The average value of the waveform = 2XAreaof1sttriangle+Areaof2ndtriangleπ**

**= 2Xπ3X12X6+6(π2–π3)π**

**= 2π+ππ = 3 Volt.**

**5. For a series RLC circuit excited by a unit step voltage, V**

_{c}is __________**a) 1 – e**

^{-t/RC}b) e

^{-t/RC}

c) e

^{t/RC}

d) 1

**Answer: a**

**Explanation: At t = 0, V**

_{c}= 0 and at t = ∞, V_{c}= 1.**This condition can be satisfied only by (1 – e**

^{-t/RC}).**6. In the circuit given below, a dc circuit fed by a current source. With respect to terminals AB, Thevenin’s voltage and Thevenin’s resistance are ____________**

a) V

_{TH}= 5 V, R

_{TH}= 0.75 Ω

**b) V**

_{TH}= 0.5 V, R_{TH}= 0.75 Ωc) V

_{TH}= 2.5 V, R

_{TH}= 1 Ω

d) V

_{TH}= 5 V, R

_{TH}= 1 Ω

**Answer: b**

**Explanation: V**

_{TH}= 1X2040 X 1 = 0.5 V**Also, R**

_{TH}= 1X3040 = 0.75 Ω.**7. In the circuit given below, the value of R is ____________**

a) 12 Ω

**b) 6 Ω**

c) 3 Ω

d) 1.5 Ω

**Answer: b**

**Explanation: The resistance of parallel combination is given by,**

**R**

_{eq}= 403 – 10 = 3.33 Ω**Or, 13.33=112+115+115**

**Or, R = 6 Ω.**

**8. A circuit consists of an excitation voltage V**

_{S}, a resistor network and a resistor R. For different values of R, the values of V and I are as given, R = ∞, V = 5 volt; R = 0, I = 2.5 A; when R = 3 Ω, the value of V is __________a) 1 V

b) 2 V

**c) 3 V**

d) 5 V

**Answer: c**

**Explanation: When R = ∞, V = 5v,**

**Then, V**

_{oc}= 5V and the circuit is open**When R = 0, I = 2.5A**

**Then, I**

_{sc}= 2.5 and the circuit is short circuited.**So, R**

_{eq}= VOCISC**= 52.5 = 2 Ω**

**Hence the voltage across 3 Ω is 3 volt.**

**9. Three inductors each 30 mH are connected in delta. The value of inductance or each arm of equivalent star is _____________**

**a) 10 mH**

b) 15 mH

c) 30 mH

d) 90 mH

**Answer: a**

**Explanation: We know that if an inductor L is connected in delta, then the equivalent star of each arm = LXLL+L+L**

**Given that, L = 30 mH**

**= 30X3030+30+30**

**= 90090 = 10 mH.**

**10. In a series RLC circuit having resistance R = 2 Ω, and excited by voltage V = 1 V, the average power is 250 mW. The phase angle between voltage and current is ___________**

a) 75°

b) 60°

c) 15°

**d) 45°**

**Answer: d**

**Explanation: VI cos θ = 0.25 or I cos θ = 0.25**

**Or, Z cosθ = 2**

**Or, VI cosθ = 2**

**Or, cos θ = 12√**

**So, from the above equations, cos θ = 0.707 and θ = 45°.**

**11. In the circuit given below, the equivalent capacitance is _________________**

a) 1.6 F

**b) 3.1 F**

c) 0.5 F

d) 4.6 F

**Answer: b**

**Explanation: C**

_{CB}= (C2C3C2+C3) + C_{5}= 7.5 F**Now, C**

_{AB}= (C1CCBC1+CCB) + C_{6}= 8 F**C**

_{XY}= CAB×C4CAB+C4 = 3.1 F.**12. In the circuit given below, the equivalent capacitance is ______________**

**a) 5.43 μF**

b) 4.23 μF

c) 3.65 μF

d) 5.50 μF

**Answer: a**

**Explanation: The 2 μF capacitor is in parallel with 1 μF capacitor and this combination is in series with 0.5 μF.**

**Hence, C**

_{1}= 0.5(2+1)0.5+2+1**= 1.53.5 = 0.43**

**Now, C**

_{1}is in parallel with the 5 μF capacitor.**∴ C**

_{EQ}= 0.43 + 5 = 5.43 μF.**13. In the circuit given below, the voltage across AB is _______________**

a) 250 V

b) 150 V

**c) 325 V**

d) 100 V

**Answer: c**

**Explanation: Loop current I**

_{1}= 5020 = 2.5 A**I**

_{2}= 10020 = 5 A**V**

_{AB}= (50) (2.5) + 100 + (5) (20)**= 125 + 100 + 100**

**= 325 V.**

**14. The number of non-planar graph of independent loop equations is ______________**

a) 8

b) 12

**c) 3**

d) 5

**Answer: c**

**Explanation: The total number of independent loop equations are given by L = B – N + 1 where,**

**L = number of loop equations**

**B = number of branches = 10**

**N = number of nodes = 8**

**∴ L = 10 – 8 + 1 = 3.**

**15. In the circuit given below, M = 20. The resonant frequency is _______________**

a) 4.1 Hz

b) 41 Hz

c) 0.41 Hz

**d) 0.041 Hz**

**Answer: d**

**Explanation: I**

_{EQ}= L_{1}+ L_{2}+ 2M**L**

_{EQ}= 10 + 20 + 2 × 120 = 30.1 H**∴ F**

_{O}= 12πLC√**= 12π30.1×0.5√**

**= 0.041 Hz.**