Home Subjects Electrical Engg (EEE)

Electrical Engg (EEE)

Electrical machines, power systems, circuits

321 Q 7 Topics Take Mock Test
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Difficulty: All Easy Medium Hard 311–320 of 321
Topics in Electrical Engg (EEE)
Q.311 Medium Circuit Analysis
In Norton's theorem, the Norton equivalent current (IN) is found by:
A Short-circuiting the load terminals and calculating the current through the short circuit
B Open-circuiting the load terminals
C Removing all independent sources
D Calculating Thévenin voltage divided by Thévenin resistance
Correct Answer:  A. Short-circuiting the load terminals and calculating the current through the short circuit
EXPLANATION

Norton current is the short-circuit current at the load terminals. IN = VTh/RTh.

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Q.312 Medium Circuit Analysis
The equivalent resistance of two resistors in parallel is always:
A Less than the smaller resistor
B Greater than both resistors
C Equal to their arithmetic mean
D Equal to their sum divided by 2
Correct Answer:  A. Less than the smaller resistor
EXPLANATION

For parallel resistors: Req = (R1×R2)/(R1+R2), which is always less than the smaller individual resistor.

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Q.313 Medium Circuit Analysis
In a series RLC circuit at resonance, the impedance is:
A Minimum and equal to R
B Maximum and equal to R
C Equal to XL + XC
D Equal to √(R² + L²)
Correct Answer:  A. Minimum and equal to R
EXPLANATION

At resonance, XL = XC, so they cancel. Z = R (minimum), and current is maximum.

Test
Q.314 Medium Circuit Analysis
In a delta-wye (Δ-Y) conversion, if the delta resistances are RA, RB, RC, the wye resistance R1 (between node 1 and star point) is:
A (RA×RB)/(RA+RB+RC)
B RA+RB+RC
C (RA×RC)/(RA+RB+RC)
D RA/3
Correct Answer:  A. (RA×RB)/(RA+RB+RC)
EXPLANATION

In Δ-Y conversion, each Y-resistor equals the product of adjacent Δ-resistors divided by the sum of all Δ-resistors.

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Q.315 Medium Circuit Analysis
The bandwidth of a resonant circuit is related to Q-factor by:
A BW = f₀/Q
B BW = Q/f₀
C BW = Q×f₀
D BW = f₀²/Q
Correct Answer:  A. BW = f₀/Q
EXPLANATION

Bandwidth is inversely proportional to Q-factor. Higher Q means narrower bandwidth. BW = f₀/Q.

Test
Q.316 Medium Circuit Analysis
A bridge circuit is balanced when:
A Z1×Z3 = Z2×Z4
B Z1+Z2 = Z3+Z4
C Z1/Z2 = Z3/Z4
D Z1-Z2 = Z3-Z4
Correct Answer:  A. Z1×Z3 = Z2×Z4
EXPLANATION

For a balanced AC bridge, the product of opposite arm impedances must be equal: Z1×Z3 = Z2×Z4.

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Q.317 Medium Circuit Analysis
A capacitor and inductor are in series with resonance occurring at 100 Hz. If L = 25 mH, the capacitance C is approximately:
A 101.3 µF
B 50.65 µF
C 202.6 µF
D 25.3 µF
Correct Answer:  A. 101.3 µF
EXPLANATION

At resonance: f = 1/(2π√LC). C = 1/(4π²f²L) = 1/(4π²×10000×0.025) ≈ 101.3 µF.

Test
Q.318 Medium Circuit Analysis
In Thevenin's theorem, the Thevenin equivalent voltage (VTh) is found by:
A Open-circuiting the load and measuring voltage across its terminals
B Short-circuiting the load and measuring current
C Applying a test voltage at the terminals
D Calculating the average voltage across all elements
Correct Answer:  A. Open-circuiting the load and measuring voltage across its terminals
EXPLANATION

Thevenin voltage is the open-circuit voltage measured across the load terminals after removing the load.

Test
Q.319 Medium Circuit Analysis
The Q-factor (quality factor) of a resonant circuit is defined as:
A Q = (1/R)√(L/C)
B Q = R√(L/C)
C Q = ω₀L/R or 1/(ω₀RC)
D Q = R/(ω₀L)
Correct Answer:  C. Q = ω₀L/R or 1/(ω₀RC)
EXPLANATION

Q-factor represents selectivity and bandwidth characteristics. For series RLC: Q = ω₀L/R = 1/(ω₀RC) = (1/R)√(L/C).

Test
Q.320 Medium Circuit Analysis
In a parallel RC circuit, the total admittance Y is given by:
A Y = G + jωC where G = 1/R
B Y = G - jωC
C Y = √(G² + ω²C²)
D Y = G/(1 + ωC)
Correct Answer:  A. Y = G + jωC where G = 1/R
EXPLANATION

In a parallel RC circuit, admittances add directly. Y = G + jBC where G = 1/R and BC = ωC.

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