An auto-transformer having 1250 turns is connected across a 250 V AC supply. What secondary voltage will be obtained if a tap is taken at 900th turn?
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Solution
V1= 250
N1= 1250
N2= 900Secondary voltage v2=v1×\(\frac{N_{2}}{N_{1}}\)
V2=250×\(\frac{900}{1250}\)=180V
Minimum number of NAND gates are required to implement the following Boolean equation, by taking inputs as A, B,C, D, E, F is ____________.
\(F=\overline{A}C+\overline{D}F+\overline{B}C+\overline{E}C\)-
Solution
\(F=\overline{A}C+\overline{D}F+\overline{B}C+\overline{E}F\)
\(F=(\overline{A}+\overline{B})C+(\overline{D}+\overline{E})F\)
\(F=(\overline{A}+\overline{B})C+(\overline{D}+\overline{E})F\)
∴ Number of NAND gates required = 5
A Q-meter with insertion resistance of 0.2 Ω is used to measure the inductance of coil. Resonance occurs at an angular frequency of 106 rad/s with acapacitor of 80 PF.The inductance of the coil is _________mH.
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Solution
A Q-meter with insertion resistance of 0.2 Wis used to measure the inductance of a coil.
ω0=106 rad/s
c=80 pf
ω0=ω0
106=\(\frac{1}{\sqrt{LC}}\)
\(=\frac{1}{\sqrt{L\times 80\times 10^{-12}}}\)
⇒L = 12.5 mH
In a star/delta connection, the turns ratio is x : 1, if primary line voltage is V and the current is I then,secondary line voltage and the line current will be
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Solution
In a star/delta connection, the turns ratio is x : 1, if primary line voltage is V and the current is I.In star/delta (Y/Δ) connection,
\(\frac{V_{1L}/\sqrt{3}}{V_{2L}}=\frac{x}{I}\Rightarrow \frac{V_{1L}}{V_{2L}}=\frac{\sqrt{3}}{I}x\)
or \(\frac{V_{2L}}{V_{1L}}=\frac{I}{\sqrt{3}x}\)
⇒ V2L=\(\frac{V_{1L}}{\sqrt{3}x}=\frac{v}{\sqrt{3}x}\)
Also,\(\frac{I_{2}L}{\sqrt{3}.1_{1L}}=\frac{x}{1}\)
⇒ I2L=\(\sqrt{3}I_{LI}x=I\sqrt{3}x\)
For multiple scattering mechanism, if individual mean free time of scattering are τm1, τm2, … Then,the effective mean free path (τm)
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Solution
The combined mobility
\(\frac{1}{\mu _{m}}=\frac{1}{\mu _{m1}}+\frac{1}{\mu _{m2}}+....\)
and \(\mu _{m}=\frac{q\tau _{m}}{m*}\)
Hence \(\frac{1}{\tau _{m}}=\frac{1}{\tau _{m1}}+\frac{1}{\tau _{m2}}+\frac{1}{\tau _{m3}}+.....\)
A coil of inductance 200 μH is magnetically coupled to another coil of inductance 800 μH. The coefficient of coupling between the coils is 0.05. The inductance if the two coils are connected in series adding mode in μH is_____________.
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Solution
L1= 200 μH
L2= 800 μH, K = 0.05
In series adding mode, inductance is given by
L = L1+ L2+ 2M
L = L1+ L2+ 2K\(\sqrt{L_{1}L_{2}}\)
= 200 + 800 + 2× 0.05 \(\sqrt{200\times 800}\)
= 1000 + 2 × 0.05 × 400= 1040 μH
The polar plot of a type-1 and order-3,open loop system is shown in the figure. The close loop system is
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Solution
The polar plot of a type-1 and order-3, open loop system.
Since, the polar plot is enclosing (–1, 0)Thus, system is unstable. This encirclement is only once hence single pole is on the right hand side of the plane.
In the Smith chart, the normalised impedance is same for
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Solution
The normalised impedance and admittance is repeated for every half wavelength of distance.
The relation exist between the circuit magnification factor Qo the resonant frequency fo and bandwidth Δ f
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Solution
\(Q=\frac{f_{o}}{\Delta f}\)
For a transform coupled power amplifier maximum theoretical efficiency will be ________ %.
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Solution
The efficiency η=50\(\frac{V_{max}-V_{min}}{V_{max}+V_{min}}\)%
max(η)=50%