Que.6(a) Using superposition theorem , find the voltage across 4 KΩ

Solution :-

Step 1 :- Consider current source 10 mA acting alone and put short circuit on 15 V & 25 V voltage sources.

Current through 4 KΩ by current division rule $latex \begin{array}{l}I=10\times10^{-3}\times\frac{1\times10^3}{4\times10^3+1\times10^3}=2\times10^{-3}A\\\\\\\\\\\\\\\\\\\\end{array}$

Step 2 :- Consider 25 V Voltage source acting alone, put open circuit on the 10 mA current source and short circuit on 15 V voltage source.

Current through 4 KΩ resistor $latex \begin{array}{l}I”=\frac VR=\frac{25}{4\times10^3+1\times10^3}=5\times10^{-3}A\\\\\\\\\\\\\\\\\\\\end{array}$

Step 3 :- Consider 15 V voltage source acting alone and put open circuit on 10 mA current source and short circuit on 25 V voltage source.

Because of short circuit , the current through the 4 KΩ resistor is 0.

∴ I’’’ = 0

Step 4 :- By applying superposition theorem,

Current through 4 KΩ resistor is

I = I’ + I’’ + I’’’

$latex \begin{array}{l}=\left(2\times10^{-3}\right)+\left(5\times10^{-3}\right)+\left(0\right)\\=7\times10^{-3}A\\\\\\\\\\\\\\\\\\\\end{array}$

Voltage across 4 KΩ resistor is,

$latex \begin{array}{l}V=IR=7\times10^{-3}\times4\times10^3\\=28V\\\\\\\\\\\\\\\\\\\\end{array}$

Que. 6(b) A series R-C circuit of R = 1000 Ω, L = 100 mH and C = 10μF . The applied voltage across the circuit is 100 V.

i) Find the resonance freq. of the ckt

ii) Find Q of the ckt at resonant freq.

iii) At what angular freq. do the half power points occur

iv) Calculate the bandwidth of the ckt.

Solution :-

Data :- R = 1000 Ω

$latex \begin{array}{l}L=100mH=100\times10^{-3}H\\c={10}_{\mu F}=10\times10^{-6}F\\\\\\\\\\\\\\\\\\\\end{array}$

i) Resonance frequency fr

$latex \begin{array}{l}f_r=\frac1{2\pi\sqrt{Lc}}\\=\frac1{2\pi\sqrt{100\times10^{-3}\times10\times10^{-6}}}\\=159.155KHZ\end{array}$

ii) Q factor of circuit at resonance frequency

$latex \begin{array}{l}Q=\frac1R\sqrt{\frac LC}=\frac1{1000}\sqrt{\frac{100\times10^{-3}}{10\times10^{-6}}}\\=0.1\\\end{array}$

iv) Band Width

$latex \begin{array}{l}Q=\frac{f_r}{B_w}\\BW=\frac{fr}Q=\frac{159.155}{0.1}=1591.55HZ\\\\\end{array}$

$latex \begin{array}{l}f_L=f_r-\frac12BW\\=159.155-\frac12\left(1591.55\right)\\=-636.62HZ\\\\\end{array}$

$latex \begin{array}{l}f_H=f_r+\frac12BW\\=159.155+\frac12\left(1591.55\right)\\=954.93HZ\\\\\\\end{array}$

Que. 6(c) Show that the total power and pf in a 3-Ф balanced system can be determined using two wattmeter method.

Solution :-

A three phase balanced voltage is applied on a balanced three phase load makes a current in each of the phases lagging by angle Φ behind the corresponding phase voltage.

Diagram

The two wattmeters must be connected in such a way that their current coils are connected in series with the two phases and their pressure coils must be connected between their respective lines and the remaining third line.

For wattmeter W1 :-

Current through current coil = IR

Potential difference across voltage coil = VRN – VYN = VRY

Phase difference between IR and VRY is (300+ Φ)

Thus, reading on wattmeter W1 is

W1= VRY IR cos(300+ Φ)

For wattmeter W2 :-

Current through current coil = IB

Potential difference across voltage coil = VBN- VYN = VBY

Phase difference between IB and VBY is (300- Φ)

Thus, reading on wattmeter W2 is

W2= VBY IB<>/sub cos(300-Φ)

Power measurement by two wattmeter method

Since the load is balanced, | IR |=| IY |=| IB |= I (Let) and

| VRY |=| VBY |= VL (Let)

∴ W1= VL I cos(300+ Φ) while W2= VL I cos(300-Φ)

Thus, the total power is given by

W= W1+ W2 = VL I cos(300+ Φ) + VL I cos(300- Φ)

= VL I [cos(300+ Φ) + cos(300- Φ) ]

$latex \begin{array}{l}=\sqrt3V_L

I\;\cos\Phi\\\\\\\\\end{array}$

$latex \begin{array}{l}\therefore\left{\cos\left(A+B\right)+\cos\left(A-B\right)=2\cos A\cos B\right}\\\\\\\\\end{array}$

Power factor (p.f ) measurement by two wattmeter method

W2- W1 = VL I sin Φ

Dividing the two equations,

$latex \begin{array}{l}=\frac{\left(W_2-W_1\right)}{\left(W_1+W_2\right)}=\frac{\tan{\displaystyle\pi}}{\sqrt3}\\\\\\\\\end{array}$

Thus,$latex \Phi=\tan-1\left[\frac{\sqrt3\left(W_2-W_1\right)}{W_1+W_2}\right]$

and power factor = cos Φ

but, power factor nature i.e lagging or leading can’t be known by this method.

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