The number of NAND gates required to construct Full Adder, Full Subtractor, Half Adder, Half Subtractor is
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Solution
Full adder:Sum =A ⊕ B ⊕ C
Carry = (A ⊕ B)C + AB
Required NAND GATES = 9Full subtractor: Difference = A ⊕ B ⊕ C
Borrow =\(\overline{A}\)B+ (A ⊕ B)C
Required NAND GATES = 9Half adder : Sum A⊕B, Carry = AB
Required NAND GATES = 5Halfsubtractor: Difference = A ⊕ B, Borrow = AB
Required NAND GATES = 5
If the field resistance of a DC shunt motor driving a constant torque load is decreased by 10% and supply voltage is also decreased by 10%, speed of motor will be approximately
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Solution
T1= T2
φ1Ia1=φ2Ia2
\(I_{f1}=\frac{V}{R_{f}}\)....(i)
\(I_{f2}=\frac{0.9V}{0.9R_{f}}=\frac{V}{R_{f}}=If_{1}\)
If1=If2
⇒φ1=φ2....(ii)
From Eqs. (i) and (ii) we get
Ia1=Ia2
V1=Kφ1H1+Ia1Ra
0.9V1=Kφ1H2+Ia1Ra
\(0.9=\frac{K\phi _{1}H_{2}+I_{a1}R_{a}}{K\phi _{1}H_{1}+I_{a1}R_{a}}\)
0.9Kφ1H1=Kφ1H2+0.1Ia1Ra
0.9H1=H2+\(\frac{0.1R_{a}I_{a1}}{K\phi _{1}}\)
Since Ra is small and K is large, H2 ≅ 0.9 H1
% change =\(\frac{H_{2}-H_{1}}{H_{1}}=\frac{0.9H_{1}_H_{1}}{H_{1}}\times 100\)
= –10% reduction in speed
Which one of the following is the correct state equation for the network shown in the given figure with x
21(t) = Vc(t) and x2(t)= iL(t) ?
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Solution
Write the network equations,
\(V(t)=R_{2}i_{L}+L_{2}\frac{di_{L}}{dt}+V_{c}\)
⇒\(x_{2}=-\frac{1}{L_{2}}x_{1}-\frac{R_{2}}{L_{2}}x_{2}+\frac{V(t)}{L_{2}}\)
\(i_{L}t=\frac{V_{c}}{R_{1}}+C_{1}\frac{dV_{c}}{dt}\)
⇒\(x_{1}=\frac{-x_{1}}{R_{1}C_{1}}+\frac{x_{2}}{C_{1}}\)
∴\(\begin{bmatrix} \dot{x_{1}}(t)\\ \dot{x_{2}}(t) \end{bmatrix}=\begin{bmatrix} \frac{-1}{R_{1}C_{2}}\, \frac{1}{C_{2}} & \\ \frac{-1}{L_{2}}\, -\frac{R_{2}}{L_{2}}& \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2} \end{bmatrix}+\begin{bmatrix} 0\\ \frac{1}{L_{1}} \end{bmatrix}V(t)\)
For a push-pull class B power amplifier whose circuit diagram is shown in figure, the efficiency at the time of maximum power dissipation is ______ .
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Solution
Maximum power dissipation occurs when,
Vm=2⁄πVCC
%\(\eta =\frac{P_{AC}}{P_{DC}}=\frac{\frac{V_{m}I_{m}}{2}}{\frac{2}{\pi }V_{CC}I_{m}}\)
=\(\frac{\frac{2}{\pi }\frac{V_{cc}I_{m}}{2}}{\frac{2}{\pi }v_{cc}I_{m}}=\frac{1}{2}=50\)%
A resistance is to be measured by the voltmeter ammeter method. The voltmeter reading is in 50 V or 100V scale and ammeter reading is in 50 mA or 100mA scale.Both the meters are guranteed for accuracy within 2% of full scale. The limit within which the resistance can be measured in Ω is_______.
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Solution
V = 50 V, I = 50 mA
\(\frac{V}{I}=\frac{50}{50\times 10^{-3}}=1k\Omega\)
Error in voltage =\(\frac{2\times 100}{50}=4\)%
Error in ammeter = \(\frac{2\times 100}{50}=4\)%
Total limiting error =4 + 4 =8%
Limiting value of resistance =\(\frac{8}{100}\times 1000=80\Omega\)
A first quadrant DC to DC chopper feeds an inductive load of 150 Ohms resistance and inductance of 52 mH. The back emf of 60 V DC from a 360 V DC source. The chopper is operated at 250 Hz with a 30% ON state dutycycle.The load average and rms voltages are
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Solution
The chopper feeds an inductive load of 150 Ohms resistance and inductance of 52 mH. The back emf of 60 V from, a 360 dc source. The chopper is operated at 250 Hz with a 30% ON state duty cycle
ON state duty cycle δ=\(\frac{30}{100}=0.3\)
Period T=\(\frac{1}{f_{s}}= \frac{1}{250Hz}=4ms\)
ON-period of the switch = 4 × 0.3= 1.2 ms
Load time constant T =\(\frac{L}{R}=\frac{52\, mh}{15\Omega }=3.466 \, ms\)
The circuit diagram can be drawn as
The load average and rms output voltage are,
Load Average voltages,\(\overline{v}_{a}=\frac{^{t}T}{T}V_{s}=\delta V_{s}=0.3\times 360=108V\)
rms output voltages,\(v_{r}=\sqrt{\frac{^{t}T}{T}}V_{s}=\sqrt{\delta }V_{s}=\sqrt{0.3}\times 360\)
= 197.18 V
For a 132 kV system sequence voltage due to a highly unbalanced load V21= 0.42 pu, V2= 0.25 pu, V23= 0.15 pu.What is per-phase pu values Va, Vb, Vac respectively?
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Solution
System voltage = 132 kV
Unbalanced load V2= 0.42 pu
V2= 0.25 pu
V3= 0.15 pu
Per-phase unit voltages
Va = V1 + V2 + V3= 0.82 ∠ 0°
Vb= α2 V1+ αV2 + V23
= (1 ∠ 24°) × 0.42 + 1 ∠ 120 × 0.25+ 0.15
= 0.24 ∠ 141.487°
Vc= αV1+α2V2+ V3
= 1∠120 +0.42 + 1 ∠ 24° × 0.25+ 0.15
= 0.24 ∠ 141.487
A 230 V,1500 rpm, 20 A separately incited DC motor is fed from 3-phase full converter. Motor armature resistance is 0.6Ω.Full converter is connected to 400 V, 50 H2 source though a delta star transformer. Motor terminal voltage is rated when converter firing angle is zero. Calculate the transformer turns ratio from primary to secondary.What is per-phase pu values Va, Vb, Vc respectively?
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Solution
Separately excited DC motor,
Voltage = 230 V, rpm= 1500
Current = 20 A
Motor armature resistance Ra= 0.6 Ω
Source voltage = 400 V
Source frequency= 50 Hz
For zero degree firing angle,motor terminal voltage is rated i.e., 230 V, therefore
∴\(\frac{3\sqrt{2}v_{1}}{\pi }cos0^{\circ}V_{1}=230v\)
V1=\(\frac{230\times \pi }{3\sqrt{2}\times 1}=170.34v\)
Here V is the line voltage. Per-phase voltage on transformer star-side is
Vph=\(=\frac{170.34}{\sqrt{3}}=98.35v\)
Per-phase voltage input to transformer delta = 400 V
∴Transformer phase turns ratio from primary to secondary
=\(\frac{400}{98.35}=4.067\)
Consider the following oscillator operating at 30° C
R= 10 kΩ C= 22 μF
L = 5 mH I= 5 mA
Determine the frequency of oscillation and the condition that must be obeyed by Rc to maintain the oscillation?Assume b to be very high.
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Solution
Oscillator operating at 30°C.
R = 10 kΩ, C= 22 μF, L = 5 = mH, I = 5 mA
\(f_{0}=\frac{1}{2\pi \sqrt{LC}}=\frac{1}{2\pi \sqrt{5\times 10^{-3}\times 22\times 10^{-6}}}\)Hz
= 480 Hz
For oscillation to start, amplifier gain must be> 1 Gain of the differential -amplifier
\(\frac{1}{2}g_{m}R_{c}=\frac{1}{2}\times \frac{R_{c}}{h_{ib}}\)
hib=\(\frac{V_{T}}{I_{EQ}}\)where, VT at 30° C (=303K)
\(\frac{0.0261}{l/2}=\frac{1.38\times 10^{-23}\times 303}{1.6\times 10^{-19}}\)
=\(\frac{0.0261}{2.5}k\Omega =10.44\Omega\)
∴\(\frac{1}{2}\times \frac{R_{c}}{10.44}>1\)
Hence, Rc > 20.88 Ω
Consider the following circuit consisting of D-flip flops
The circuit generates a sequence. Initially all flip-flops are cleared. The generated sequence is
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Solution
The sequence generated is 11010. It actually starts after a few clock periods.