For the circuit shown below,the Thevenin equivalent is given by
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Solution
Shorting terminal T-H
\(I_{SC}=\frac{\mu V_{F}}{R_{F}R_{o}}(R_{F}+R_{o})\times \frac{R_{F}}{R_{F}+R_{o}}=\frac{\mu V_{F}}{R_{o}}\, and\, V_{OC}=\mu V_{F}\)
So, RTH=\(\frac{V_{OC}}{I_{SC}}=\frac{\mu V_{F}}{\mu V_{F}}+R_{o}=R_{o}\)
also Vs= VF+ μVF= (μ+ 1)VF and VTH= μVF
So, VTH=\(\frac{\mu }{1+\mu }V_{s}\)
Consider the Bode magnitude plot shown in figure. Transfer function is
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Solution
Transfer function =\(\frac{ks^{2}}{(1+2s)(1+s)(1+0.2s)}\)
slope of AB=\(\frac{y_{B}-y_{A}}{x_{B}-x_{A}}=20\)
\(=\frac{32-y_{A}}{log1-log0.5}=2.0\)
⇒yA= 26 dB
Slope of OA=\(\frac{y_{A}-y_{o}}{x_{A}-x_{o}}=40\)
\(=\frac{26-y_{o}}{log0.5-log0.1}=40\)
\(y_{o}=-2db\)
So,y = mx + c
–2 = 40 log 0.1 + 20 log k
k = 79.43
A 4-pole, 3-phase, star connected slip-ring induction motor is to operate from 400 V,3-phase, 50 Hz mains.Its rotor has a standstill leakage impedance of 0.4 + j 2 ohms per phase.For negligible stator impedance and rotational losses, the resistance to be included in the rotor circuit to develop 0.8 of maximum torque at start in Ohm is _______.
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Solution
Maximum Torque, Tmax=\(\frac{180}{2\pi N_{s}}\times \frac{E_{2}^{2}}{2X_{2}}\)
NS=\(\frac{120\times 50}{4}=1500\)
∴ Tmx=\(\frac{180}{2\pi \times 1500}\times \frac{\left ( \frac{400}{\sqrt{3}} \right )^{2}}{2\times 2}\)
(∵Stator impedance is neglected E2 = V1)
\(\frac{T_{st}}{T_{max}}=\frac{2S_{m}}{S_{m}^{2}+1}\)
⇒0.8=\(\frac{2S_{m}}{S_{m}^{2}+1}\)
⇒\(0.8S_{m}^{2}-2S_{m}+0.8=0\)
⇒Sm= 2(not valid)
⇒Sm= 0.5
At maximum torque condition
\(\frac{R_{total}}{X_{2}}=S_{m}\)
\(\frac{r_{2}+r_{ext}}{X_{2}}=0.5\)
⇒0.4 + rext= 0.5 × 2
⇒rext= 0.6Ω
Ten thyristors are used in a string to with stand a DC voltage of Vs= 15kV. The maximum leakage current and recovery charge difference of thyristor are 10 mA and 150 μC,respectively. Each thyristor has a voltage-sharing resistance of R = 545 kW and capacitance of C1= 0.5mF. The maximum steady state voltage-sharing VDS(max)in volt is _______.
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Solution
nS= 10,VS= 15kV
ΔID= ΔID2= 10 mA
and ΔQ = Q2= 150 μC
R =56 kΩ
\(V_{DS_{(max)}}=\frac{v_{s}+(n_{s}-1)R/D_{2}}{n_{s}}\)
\(V_{DS_{(max)}}=\frac{1500+(10-1)\times 56\times 10^{3}\times 10\times 10^{-3}}{10}\)
\(V_{DS_{(max)}}=2004\, V\)
What is the condition at which voltage across capacitor or inductor is maximum?
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Solution
\(1=\frac{V}{\sqrt{R^{2}+\left ( \omega L-\frac{1}{\omega C} \right )^{2}}}\)
VL= l × ωL
\(V_{L}=V\times \frac{\omega L}{\sqrt{R^{2}+\left ( \omega L-\frac{1}{\omega C} \right )^{2}}}\)
\(\frac{dV_{L}}{df}=0\)
For maximum voltage across inductor
\(f_{L}=\frac{1}{2\pi \sqrt{LC}}\sqrt{1-\frac{\frac{1}{R^{2}C}}{2L}}\)
A signal whose spectrum is shown figure below is applied to the following system.
where H(f) = 5000\(\sum_{k=\infty }^{\infty}\)δ(f–5000k)and cutoff frequency of the ideal LP-filter is fc= 5 kHz.
Which of the following frequency component will be present in Y(f)
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Solution
Output of the first filter
H' (f) =X(f)* H(f)
X(f)=1⁄2[δ(f-2)+δ](f+2)
H'(f) =1⁄2[δ(f-2)+δ](f+2)*5000(f-5k)
=2500[δ (f – 5k – 2) + δ (f – 5k+ 2)
So H' (f) will contain frequency components as f = (5k ± 2)kHzwhere k = 0 to ∞.
After passing through LP-filter with cutoff frequency fc= 5 kHz spectrum of Y(f) will have component of frequencies f1 = 2 kHz, f2= 3 kHz only.
The step size of the DAC of the given figure can be changed by changing the value of Rf.
Which of the following represents the required value of Rf for a step size of 0.5 V?
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Solution
Since step size =Rf×\(\left ( \frac{5v}{8k\Omega } \right )\)
⇒ Rf=\(\frac{0.5\times 8\times 10^{3}}{5}=800\Omega\)
A five-bit DAC produces 10 mA for adigital input of 10100.For digital input of 11101 output current in mA is _____
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Solution
The digital input 101002 is equal to decimal 20. Since
IOUT= 10 mA for 20, the proportionality factor must be 0.5mA. Thus,we can find IOUT for any digit input such as 111012= 2910 as follows.IOUT= (0.5 mA)×29 = 14.5 mA
Consider the following 8085 assembly program:
MVI B, 89H
MOV A, B
MOV C, A
MVI D, 37H
OUT PORT1
HLT
The output at PORT 1 is __________
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Solution
MVI B, 89H ;89→ B
MOV A, B ;B → A
MOV C, A ;A → C
MVI D, 37 H ;37 →D
OUT PORT1 ;Display A
The contents of A is 89 H.
Assuming that the diodes in the given circuit are ideal, the voltage V0 is __________
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Solution
Given circuit is,
We can observe that diode D2 is always off, whether D1, is ON or OFF. So equivalent circuit is.
D1 is ON in this condition
\(V_{0}=\frac{10}{10+10}\times 10=5\, volt\)