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৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]

পরীক্ষা৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]তারিখতারিখ অনির্ধারিতসময়19 minutes
মোট প্রশ্ন৩০
সিলেবাস
Exam 2 Network Theorems (DC & AC): Superposition Theorem, Thevenin’s Theorem, Norton’s Theorem, Maximum Power Transfer Theorem RC, RL and RLC Circuits: RC and RL High-Pass, Low-Pass Filters, Band-pass, Series and Parallel RLC Resonances AC Power and Polyphase Circuits: Apparent Power, Power Factor and Complex Power, Polyphase Systems, Single-Phase Three-Wire Systems, Three-Phase Y-Y Connection, The Delta Connection, Power measurement in Three-Phase Systems [Source: Classes 1–2 and relevant books]
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৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২]

৪৯তম বিসিএস ⎯ তথ্য ও যোগাযোগ প্রযুক্তি (EEE) [ ৮৯২] · তারিখ অনির্ধারিত · ৩০ প্রশ্ন

.
What type of filter allows frequencies below a cutoff frequency to pass through?
  1. High-pass filter
  2. band-pass filter
  3. low-pass filter
  4. band-stop filter
ব্যাখ্যা

A low-pass filter (LPF) attenuates frequencies above its cutoff frequency while passing lower frequencies unchanged

.
In a passive RC high-pass filter, the output is taken across which component?
  1. resistor
  2. capacitor
  3. inductor
  4. diode
ব্যাখ্যা

In an RC high-pass filter, the capacitor blocks low frequencies, while higher frequencies pass through to the resistor. Thus, the output voltage is measured across the resistor.  

.
What is the primary function of a band-stop (notch) filter?
  1. Allow all frequencies to pass
  2. Block a specific range of frequencies
  3. Only allow very high frequencies
  4. Only allow very low frequencies
ব্যাখ্যা

Band-stop filters attenuate a narrow frequency band  while passing others.  

.
The quality factor (Q) of a band-pass filter indicates
  1. The bandwidth of the filter
  2. The sharpness of the resonance peak
  3. The gain of the filter
  4. The phase shift of the filter
ব্যাখ্যা

Higher Q means a narrower bandwidth  

.
In a series RLC circuit at resonance, the impedance is-
  1. Maximum
  2. Minimum
  3. Zero
  4. Equal to XL
ব্যাখ্যা

At resonance, XL=XC, so the net reactance is zero, making impedance purely resistive (Z=R), which is the minimum possible.

.
At resonance in a series RLC circuit, the phase angle between voltage and current is-
  1. 90°
  2. 45°
  3. 180°
ব্যাখ্যা

Since XL=XC, the circuit is purely resistive, so voltage and current are in phase.

.
At resonance, the voltage across the inductor (VL) and capacitor (VC) in a series RLC circuit are-
  1. Equal and in phase
  2. Equal and 180° out of phase
  3. Unequal and 90° out of phase
  4. Zero
ব্যাখ্যা

VL=IXL  and VC =IXC  are equal in magnitude but cancel each other due to a 180° phase difference.

.
If the frequency increases above resonance in a series RLC circuit, the circuit becomes-
  1. Capacitive
  2. Inductive
  3. Resistive
  4. Remains resonant
ব্যাখ্যা

At higher frequencies, XL>XC making the circuit inductive.

.
In a series RLC circuit, if R is increased, the Q-factor will-
  1. Increase
  2. Decrease
  3. Remain same
  4. Become zero
ব্যাখ্যা

Q=XL/R so increasing R reduces Q.

১০.
At resonance in a parallel RLC circuit, the line current is-
  1. Maximum
  2. Minimum
  3. Zero
  4. Equal to IL or IC
ব্যাখ্যা

 The reactive branch currents cancel, minimizing the total line current.

১১.
A parallel RLC circuit is often called -
  1. Rejector circuit
  2. Acceptor circuit
  3. Tank circuit
  4. Filter circuit
ব্যাখ্যা

It stores energy alternately between L and C, resembling a tank.

১২.
The half-power frequencies are those where the power is-
  1. Maximum
  2. 50% of maximum
  3. 70.7% of maximum voltage
  4. Zero
ব্যাখ্যা

Half-power points occur at 1/√2 of peak current/voltage.

১৩.
Superposition in AC circuits requires analyzing the circuit separately for
  1. Magnitude only
  2. Frequency only
  3. Both magnitude and phase
  4. Only resistive components
ব্যাখ্যা

Both magnitude and phase

১৪.
Superposition cannot be used to calculate-
  1. Voltage across a resistor
  2. Power dissipated
  3. Current through a capacitor
  4. Equivalent impedance
ব্যাখ্যা

Power is nonlinear (P=I2R or V2/R), so superposition doesn’t apply.

১৫.
For a purely resistive AC circuit, max power occurs when-
  1. RL =Rth
  2. RL =0
  3. RL =∞
  4. RL =2Rth
ব্যাখ্যা

Resistive loads simplify to matching resistances.

১৬.
For a circuit containing a dependent current source (2Iₓ) and independent voltage source (10V), superposition-
  1. Can be applied normally
  2. Requires keeping the dependent source active always
  3. Only works if dependent source is linearized
  4. Cannot be used at all
ব্যাখ্যা

Dependent sources must remain active during superposition analysis as they rely on other circuit variables.

১৭.
In a network where Isc = (2-j3)A and Voc = (10+j15)V, the Norton admittance YN is-
  1. (0.2-j0.3)S
  2. (0.15+j0.1)S
  3. (2-j3)S
  4. (10+j15)S
ব্যাখ্যা

YN = I_sc/V_oc = (2-j3)/(10+j15) = 0.15+j0.1S 

১৮.
A circuit draws 10A RMS at 120V with PF=0.8PF=0.8 lagging. The real power is-
  1. 460W
  2. 1200W
  3. 800W
  4. 960W
ব্যাখ্যা

P=VI×PF=120×10×0.8=960WP=VI×PF=120×10×0.8=960W.

১৯.
The power factor angle for P=3kW and S=5kVA is
  1. 53.13°
  2. 36.87°
  3. 60°
  4. 45°
ব্যাখ্যা

θ=cos⁡−1(P/S)=cos⁡−1(0.6)=53.13°

২০.
For a load with S=100∠45°S=100∠45° VA, the real and reactive powers are-
  1. P=70.7W Q=70.7VAR
  2. P=100W Q=70VAR
  3. P=50W Q=86.6VAR
  4. P=0W Q=100VAR
ব্যাখ্যা

P=Scos⁡(45°)=70.7W,  Q=Ssin⁡(45°)=70.7VAR

২১.
A circuit has P=800W and Q=600VAR. Its power factor is-
  1. 0.6 lagging
  2. 0.8 lagging
  3. 0.6 leading
  4. 0.8 leading
ব্যাখ্যা

S=√(P2 +Q2 )=1000VA, PF=P∣S∣=0.8, Since Q>0Q>0, the PF is lagging (inductive load).

২২.
In the two wattmeter method, if one wattmeter reads zero, the power factor of the load is-
  1. 0.5 lagging
  2. 0.866 lagging
  3. 0.707 leading
  4. 0
ব্যাখ্যা

If one wattmeter reads zero, the phase angle ϕ=60°
Power factor cos⁡ϕ=cos⁡60°=0.5
Since the reading becomes zero due to a lagging phase angle, the power factor is 0.5 lagging.

২৩.
In the two wattmeter method, if the load is purely inductive, what will be the readings of the two wattmeters?
  1. Both positive and equal
  2. One positive and one negative with equal magnitude
  3. Both zero
  4. One zero and one positive
ব্যাখ্যা

one wattmeter reads positive and the other negative with the same magnitude.

২৪.
In a balanced 3-phase system, the two wattmeter readings are 1000 W and 500 W, respectively. What is the reactive power (Q) of the system?
  1. 1500√3 VAR
  2. 750 VAR
  3. 866 VAR
  4. 500√3 VAR
ব্যাখ্যা

Q=√3(W1−W2)=500√3

২৫.
In the two wattmeter method, if the phase sequence is reversed, what happens to the readings?
  1. Both readings remain the same
  2. The readings interchange
  3. One becomes zero
  4. Both become negative
ব্যাখ্যা

Swapping Vab and Vcb causes the readings to interchange 

২৬.
4. Why is Wye-Wye (Y-Y) connection rarely used in high-voltage transmission?
  1. It cannot step up/down voltage
  2. It causes a 180° phase reversal
  3. It suffers from neutral instability and third-harmonic issues 
  4. It is too expensive
ব্যাখ্যা

Unbalanced loads cause voltage fluctuations.  Requires a tertiary winding or grounding for practical use.

২৭.
A Delta-connected load has a line current of 50 A. What is its phase current (\(I_P\)) under balanced conditions?
  1. 50 A 
  2. 28.9 A 
  3.  86.6 A 
  4.  25 A
ব্যাখ্যা

IL=√3Ip

২৮.
A polyphase system is generated by __
  1. Having two or more generator windings separated by equal electrical angle.
  2. Having generator windings at equal distances 
  3.  None of the above
  4. A and C
ব্যাখ্যা

A generator having two or more electrical windings which are separated by equal electrical angle generates a polyphase electrical system.

২৯.
For a star connection network, consuming power of 1.8kW and power factor 0.5, supply voltage of 230 Volts. phase impedance=?
  1. 10.6
  2. 25.8
  3. 20.5
  4. 14.7
ব্যাখ্যা

Line voltage, VL = 230 V
Power Factor, cosφ = 0.5
Power consumed = P = 1800 Watts = √3 VL x IL x cosφ
 line current, IL = 9 Amperes
Since it is a star connection, phase current = line current = 9 Amperes
Phase Voltage, Vph = VL/√3 = 132.8 Volts
Phase Impedance, Zph = Vph/Iph = 14.7 Ohms

৩০.
the average value of a sinusoidal voltage source over one period of time is- 
  1. 0.637Vm
  2. 0.707Vm
  3. 1.11Vm
  4. 0
ব্যাখ্যা

for sine or cosine function, over one period of time , average value is zero