A two element series circuit consumes 700 V of power and has power factor of 0.707 leading when energized by a voltage source of waveform \(v=141\sin\left(314t+30^{\circ}\right)\). Find out the circuit element.
The power dissipated by the coil is \(P_C=I_{rms}V_{rms}\ cos\ \phi\)
For this problem \(V_{rms}=\frac{V_{peak}}{\sqrt2}=\frac{141}{\sqrt2}=100\ V\)
Given that
\(I_{rms}V_{rms}\ cos\ \phi=700\)
\(I_{rms}=\frac{700}{V_{rms}\ cos\ \phi}=\frac{700}{100\times0.707}=7\sqrt2\ Amp\).
Impedance of the coil \(Z_c=\frac{V_{rms}}{I_{rms}}=\frac{100}{7\sqrt2}=10.04\ \mathrm{\Omega}\)
The power factor of the coil is 0.707 leading which means that the coil has a resistor and a capacitor.
\(\phi=\cos^{-1}{0.707}={45}^0\)
Resistance \(R=10.04\ cos\ {45}^0=10.04\times0.707=7.1\ \mathrm{\Omega}\)
\(X_C=\frac{1}{\omega C}=10.04\times sin\ {45}^0=10.04\times0.707=7.1\ \mathrm{\Omega}\)
\(C=\frac{1}{\omega X_C}=\frac{1}{2\pi f\times7.1}=0.45\ mF\)
So the circuit elements are a 7.1 \(\mathrm{\Omega}) resistor and 0.45 mF capacitor.
Explain briefly the VSB-SC modulation
Find the Fourier transform of the following:
Fig. 1
Write short note on LVDT
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Fig.22(a)