For a supernode with Va−Vb = 5 V and Va/2 + Vb/3 = 0 A, find Va.
A −3 V
B 2 V
C 3 V
D 5 V
Answer: B. 2 V. The constraint gives Va = Vb+5; substitution into KCL yields Vb = −3 V and Va = 2 V.
A supermesh has I1−I2 = 2 A and 2I1+3I2 = 9 V. Find I1.
A 3 A
B 1 A
C 2 A
D 5 A
Answer: A. 3 A. Using I1 = I2+2 in the second equation gives I2 = 1 A, hence I1 = 3 A.
A network contains dependent sources and all independent sources are zeroed. How can its port resistance be found?
A Apply a test source
B Discard controlled sources
C Short the output permanently
D Measure open voltage only
Answer: A. Apply a test source. A test voltage or current at the port reveals R = Vtest/Itest while dependent sources remain active.
For a reciprocal two-port described by z parameters, which equality holds?
A z11 = z12
B z11 = z22
C z12 = z21
D z12 = −z21
Answer: C. z12 = z21. Reciprocity requires equal transfer impedances in the two directions.
For a symmetric two-port described by z parameters, which equality holds?
A z11 = z22
B z12 = 0
C z21 = 0
D z11 = −z22
Answer: A. z11 = z22. Interchanging identical ports leaves their driving-point impedances equal.
Under the standard ABCD convention with output current directed outward, what determinant indicates reciprocity?
A AD−BC = 0
B AD−BC = 1
C A+D = 1
D BC = 1
Answer: B. AD−BC = 1. A reciprocal two-port has unit transmission-matrix determinant.
A two-port has z11=5 Ω, z12=z21=2 Ω and z22=4 Ω. With a 4 Ω load, find input impedance.
A 3 Ω
B 4.5 Ω
C 5 Ω
D 6 Ω
Answer: B. 4.5 Ω. Zin = z11−z12z21/(z22+RL) = 5−4/8 = 4.5 Ω.
A delta has RAB=3 Ω, RBC=6 Ω and RCA=9 Ω. Find the equivalent wye arm at A.
A 2 Ω
B 3 Ω
C 1.5 Ω
D 4.5 Ω
Answer: C. 1.5 Ω. RA = RAB×RCA/(RAB+RBC+RCA) = 27/18 = 1.5 Ω.
Wye arms at A, B and C are 1 Ω, 2 Ω and 3 Ω. What delta resistance connects A and B?
A 3 Ω
B 6 Ω
C 11/3 Ω
D 11 Ω
Answer: C. 11/3 Ω. RAB = RA+RB+(RA×RB)/RC = 1+2+2/3 = 11/3 Ω.
Across 12 V, bridge dividers have top/bottom resistors 2/4 Ω and 3/3 Ω. Their open midpoints differ by how much?
A 0 V
B 2 V
C 4 V
D 6 V
Answer: B. 2 V. The midpoint potentials are 8 V and 6 V relative to the lower rail.
A 5 Ω resistor joins nodes 1 and 2. What is its off-diagonal nodal-admittance entry Y12?
A 0.2 S
B −5 S
C 5 S
D −0.2 S
Answer: D. −0.2 S. A shared resistor contributes minus its conductance, −1/5 S, off the diagonal.
Node 1 connects to ground by 2 Ω and to node 2 by 4 Ω. Find Y11.
A 0.25 S
B 0.5 S
C 1.5 S
D 0.75 S
Answer: D. 0.75 S. The diagonal sums connected conductances: 1/2+1/4 = 0.75 S.
A first-order RC voltage starts at 2 V and tends to 10 V. At one time constant, it is about what?
A 2.94 V
B 5.00 V
C 9.26 V
D 7.06 V
Answer: D. 7.06 V. v(τ) = 10+(2−10)e^(−1) ≈ 7.06 V.
An RL current starts at 1 A and tends to 5 A. At one time constant, it is about what?
A 1.47 A
B 3.53 A
C 2.00 A
D 4.63 A
Answer: B. 3.53 A. i(τ) = 5+(1−5)e^(−1) ≈ 3.53 A.
For an AC source with Zth = 4+j3 Ω, which complex load gives maximum power?
A 4+j3 Ω
B 3−j4 Ω
C 5 Ω
D 4−j3 Ω
Answer: D. 4−j3 Ω. Conjugate matching chooses the complex conjugate of the source impedance.
If the AC source has Zth = 6+j8 Ω and the load must be purely resistive, which load maximizes power?
A 6 Ω
B 8 Ω
C 10 Ω
D 14 Ω
Answer: C. 10 Ω. For a purely resistive load, the optimum is |Zth| = √(6²+8²) = 10 Ω.
A 10 V rms source drives an impedance of 3+j4 Ω. What is current magnitude?
A 2 A
B 1 A
C 2.5 A
D 5 A
Answer: A. 2 A. The impedance magnitude is 5 Ω, so 10/5 = 2 A rms.
A spanning tree of a connected graph with six nodes contains how many branches?
A Four
B Six
C Five
D Seven
Answer: C. Five. Any spanning tree of a connected n-node graph has n−1 branches.
Mesh equations are 5I1−2I2=10 and −2I1+4I2=0. What is I1?
A 1.25 A
B 2.5 A
C 2 A
D 5 A
Answer: B. 2.5 A. The second equation gives I2 = I1/2; then 4I1 = 10.
Nodal equations are 3Va−Vb=11 and −Va+2Vb=3. Find Vb.
A 4 V
B 3 V
C 5 V
D 7 V
Answer: A. 4 V. Solving the simultaneous equations gives Va = 5 V and Vb = 4 V.
A controlled 2Vx volt source has 4 Ω in series. What is its Norton source current?
A 2Vx A
B 4Vx A
C 0.5Vx A
D 8Vx A
Answer: C. 0.5Vx A. Source transformation gives In = (2Vx)/4 = 0.5Vx A with the control preserved.
Why does an unreferenced nodal-admittance matrix of a connected network have a dependent equation?
A One reference is missing
B No branches exist
C Every source is nonlinear
D All resistors vanish
Answer: A. One reference is missing. KCL equations sum to zero; fixing one node as reference removes the redundant degree of freedom.
A 2 F capacitor has initial voltage 4 V. With current entering its positive terminal, what is I(s)?
A 2sV(s)−8
B 2sV(s)+8
C V(s)/(2s)
D 8sV(s)
Answer: A. 2sV(s)−8. Laplace transforming i=C dv/dt gives I(s)=CsV(s)−Cv(0−)=2sV(s)−8.
A ladder has 2 Ω in series with 3 Ω parallel to a 6 Ω + 3 Ω branch. Find its input resistance.
A 2.25 Ω
B 5 Ω
C 11 Ω
D 4.25 Ω
Answer: D. 4.25 Ω. The inner series branch is 9 Ω; 3 Ω parallel 9 Ω is 2.25 Ω, then add 2 Ω.
At matched load resistance in a resistive Thévenin circuit, what fraction of source power reaches the load?
A 25%
B 75%
C 100%
D 50%
Answer: D. 50%. Equal source and load resistances dissipate equal power.