The following program is written for 8085 microprocessor to add two byte located at memory addresses 1FEE and 1FFF
LXI H, 1FF
MOV B, M
INR L
MOV A, M
ADD B
INR L
MOV M, A
XOR A
On completion of the execution of the program, the result of addition is found
The circuit shown in figure, the output V follows an equation \(\frac{d^{2}V}{dt}+a\frac{dV}{dt}\)+bv= f(t). Then (a, b, f (t)) would be
-
Solution
Output of op-amp(3) is V. Op-amp is working as integralor with given RC = 1. So, the input of op-amp (3) is-\(\frac{dV}{dt}\).
The output of op-amp (2) is \(\frac{dV}{dt}\)op-amp (2) is working as integrator. Hence, the input of op-amp (2) is \(\frac{dV^{2}}{dt^{2}}\).
The output of op-amp (1) is \(\frac{dV^{2}}{dt^{2}}\).
The op-amp (1) is working as non inverting adder
\(\frac{dV^{2}}{dt^{2}}=\left ( 1+\frac{R}{R} \right )\left ( \frac{R}{2R} \right )\left ( -\frac{dV}{dt}+e^{t} \right )\)
\(\frac{dV^{2}}{dt^{2}}=-\frac{dV}{dt}+e^{t}\)
\(\frac{dV^{2}}{dt^{2}}+\frac{dV}{dt}=e^{t}\)
Comparing with the given expression
\(\frac{dV^{2}}{dt^{2}}+\frac{adV}{dt}+bv=f(t)\)
a=1
b=0
f (t)=e
The overall transfer function of the system in figure is
-
Solution
We can write from block diagram
Y1=U+ X2 H
Y2=U+ X1 H
and X1=GY1= G (U+ X2 H) = GU+ GHX2
X2=GY2= G (U+ X1 H) =GU+ GX1 H
Y=X1+ X2= GU+ GH X2+ GU+ GHX1
Y=2GU+ GH (X1+ X2)
Y=2GU+ 2GHYY (1 – 2GH) = 2GU
Transfer function = Y/U=\(\frac{2G}{1-GH}\)
For a stable closed loop system, the gain at phase cross-over frequency should always be
-
Solution
Gain at ωpc should be less than 1
So,GM > 0 dB
\(GM=\frac{1}{M}\mid\omega =\omega _{pc}\)
The industance of a 20A electrodynamic ammeter changes uniformly at the rate of 0.0035mH/degree. The spring constant is 10–6N-m/degree.The angular deflection at full scale in degree is ________.
-
Solution
Converting \(\frac{dM}{d\Theta }\) into H/rad to get result in degree.
\(\frac{dM}{d\Theta }=\frac{0.0035\times 10^{-6}}{\pi /180}=0.2\times 10^{-6}\)H/rad
∴ Deflection θ=\(\frac{20^{2}}{10^{-6}}\times 0.2\times 10^{-6}\)
θ= 80°
A 4 bit ripple counter is made using flip-flop having apropagation delay of 10 ns each. The worst case delay of ripple counter will be
-
Solution
In ripple counter, the clock pulse will go one to another in each flip-flop.the propagation of one flip-flop is 10ns. The worst delay of ripple counter will be
=4×10ns= 40 ns
A 200/100V, single-phase transformer is rated 10 kVA. This transformer is connected as an auto-transformer across a 50 V supply (the winding designed for 100 V should be connected across 50 V supply) for getting maximum output voltage. If the load current is now 8A, then the input current of the auto-transformer in ampere is ________
-
Solution
For 1 – φ transformer, \(\frac{N_{2}}{N_{1}}=\frac{100}{200}=\frac{1}{2}\)
For this single phase transformer if we give 50V supply for the winding designed for 100V. (i.e., low voltage side) then secondary voltage of 1 – φ transformer is
2⁄1×500=100V
Then equivalent auto transformer for getting maximum out put voltage is as shown below. For getting maximum out put voltage two windings should be connected with series aiding polarities.
V2= 100V + 50V = 150V
Given load current = 8A
∴Output VA = 150 × 8 1200 VA
Input current to the auto transformer is \(\frac{1200}{50}\)1200=24A
The primary and secondary windings of a 40 kVA, 6600/250V single phase transformer have resistance of 10W ad 0.04W respectively. The total leakage reactance is 30 W as referred to the primary winding, the half load regulation at power factor of 0.8 lagging in percent is ________.
-
Solution
V1= 6600 V, V2= 250 V
R1= 10 Ω, R2= 0.04 Ω
XP =10 Ω and cos φ = 0.8
\(K=\frac{V_{2}}{V_{1}}=\frac{250}{6600}=0.0378\)
I2=\(\frac{40\times 10^{3}}{250}=160A\)
Referred to secondary side.
RS= k2 RS + RS
RS= (0.0378)2× 10 + 0.04 = 0.0542 Ω
Xs= k2 Xp= 0.0429 Ω
Full load regulation
=\(\left [ \frac{I_{2}R_{s}cos\phi +I_{2}X_{s}sin\phi }{E_{2}} \right ]\times 100\)
=\(\frac{50[0.542\times 0.8+0.0429\times 0.6]}{250}\times 100\)
= 2.2%
The distribution function Fx (x) of a random variable x is shown in figure. The probability that x= 4 is
-
Solution
At x= 4, there is a jump of value from 0.40 to 0.50.
Probability of event at x = 4 is
P (x=4) =Fx (x = 4+) –Fx (x = 4-)
=0.50 – 0.40 = 0.10
The Nyquist plot shown of the system is
-
Solution
From the Nyquist-plot shown above have –ve gain margin.
Hence, the system is unstable.