In a network, when using the cut-set method, a cut separates the network into two parts. How many equations are needed for a network with 'b' branches and 'n' nodes to find all branch currents?
Answer: B
Number of independent loop equations needed = b - n + 1 (where b is branches and n is nodes). This represents the circuit's cyclomatic complexity.
Q.5Hard
For a passive linear network, the input impedance looking into terminals is Z(s). If a unit impulse is applied as input, the impulse response h(t) is the:
Answer: C
Impulse response h(t) = L⁻¹[Y(s)] where Y(s) is admittance. For impedance Z(s), admittance Y(s) = 1/Z(s).
Advertisement
Q.6Hard
For a two-port network, when port 2 is open-circuited (I₂=0), the parameter measured is:
Answer: A
In hybrid (h) parameters: h₁₁ is input impedance with output open (I₂=0). h₁₁ = V₁/I₁|I₂=₀ measured in ohms.
Q.7Hard
The cut-set matrix is used in circuit analysis to:
Answer: B
Cut-set matrix provides a systematic way to write KVL equations for fundamental cut-sets, an alternative to mesh analysis.
Q.8Hard
In a circuit with mutual inductance M between two coils, if the coupling coefficient k = 0.8, L1 = 10H, then L2 is at least:
Answer: A
Coupling coefficient k = M/√(L1·L2). For k=0.8 and L1=10H: 0.8 = M/√(10·L2). Since M ≤ √(L1·L2), we get L2 ≥ (M/k)²/L1 = (10·k²)/k² = L1/k² = 010.64 = 15.625H minimum requires M evaluation, but using k = M/√(L1L2) ≤ 1 gives L2 ≥ 6.25H.
Q.9Hard
A circuit has a loop containing 5V source, 10V source (opposite polarity), 2Ω, and 3Ω resistors. Using KVL, the current magnitude in the loop is:
Answer: A
Applying KVL around loop: 5 - 10 + 2I + 3I = 0 (taking voltages in one direction). This gives -5 + 5I = 0, so I = 1A.
Q.10Hard
In ABCD parameters of a two-port network, parameter C represents:
Answer: B
In ABCD parameters: V1 = AV2 - BI2; I1 = CV2 - DI2. C = I1/V2 (with I2=0) is the forward transadmittance with dimensions of admittance (Siemens).
