Question

The power flow in a 3-phase, 3-wire balanced delta connected load is measured by the two wattmeter method. The reading of wattmeter 'A' is 6000 W and on wattmeter 'B' is -1000 W. If the voltage of the circuit is 440 V, 50 Hz. What is the value of capacitance connected in delta at the source to cause the whole of the power measured by wattmeter A only?

• A. $272\mu F$
• B. $251\mu F$
• C. $224\mu F$
• D. $312\mu F$

Answer 1

The correct option is A $272\mu F$

Total power in the load circuit
$P=W_1+W_2$

$=6000-1000=5000W$

$\varphi =ta{n}^{-1}\left[\frac{\sqrt{3}(W_1-W_2)}{W_1+W_2}\right]$

$=ta{n}^{-1}\left[\frac{\sqrt{3}\left[\right(6000)-(-1000\left)\right]}{6000-1000}\right]$

$=ta{n}^{-1}\left[\frac{\sqrt{3}\left(7000\right)}{5000}\right]={67.58}^o$

$cos\theta =0.381$

Load current per phase,

$I_P=\frac{5000}{\sqrt{3}\times 440\times 0.381}$

$I_P=17.22A$

$Z_P=\frac{V_P}{I_P}=\frac{440}{\left(\frac{17.22}{\sqrt{3}}\right)}=44.25\Omega$

Load resistance per phase,

$R_p=Z_{p}cos\varphi =44.25cos\left(67.58\right)$

$=44.25\times 0.381$

$R_p=16.87\Omega$

Load reactance per phase,

$X_p=Z_{p}sin\varphi =44.25sin\left(67.58\right)$

$X_p=40.9\Omega$

Reading of wattmeter B will be zero when power factor,

$cos{\varphi }^{\prime }=0.5$

${\varphi }^{\prime }={60}^o$

Since there is no change in resistance, reactance in circuit per phase,

$X_p{}^{\prime }=R_{p}tan{\varphi }^{\prime }$

$X_p{}^{\prime }=16.87\times tan{60}^o=29.22\Omega$

$X_C=X_p-X_p{}^{\prime }$

$=40.9-29.22=11.68\Omega$

$\therefore C=\frac{1}{2\pi f{X}_C}=\frac{1}{2\pi \times 50\times 11.68}$

$C=272.52\mu F$