Question

The figure shows a 3-phase load supplied from a 100 V, 50 Hz, 3-phase balanced source. The pressure coil (PC) and current coil (CC) of a wattmeter are connected to the load as shown, with the coil polarities suitably selected to ensure a positive deflection. The wattmeter reading will be _________kW.

(Assume RYB Phase sequence). Answer upto two decimal places.

• A. 86.6
• B. 8.66
• C. 866
• D. 0.86

Answer 1

The correct option is B 8.66

By converting the Star connected capacitors into Delta.
$I_{RY}=\frac{V_{RY}}{Z}=\frac{100\angle 0^o}{\frac{j1\times (-3j)}{j-3j}}=\frac{100}{\frac{3}{-2j}}=\frac{-200}{3}j$

$\therefore I_{j1\Omega }=I_{RY}\times \frac{-3j}{-3j+j}=\frac{-200}{3}j\times \frac{-3j}{-2j}$

$I_{j1\Omega }=\frac{-200}{3}j\times \frac{3}{2}=-100j$

$\therefore I_{j1\Omega }=100\angle -{90}^o$

$\therefore$ Current through the current coil
$=100\angle -{90}^o$

Voltage across potential coil is $V_{YB}$

$\therefore V_{YB}=100\angle -{120}^o$

$\therefore$ Wattmeter reading
$=100\times 100cos{30}^o=8.66kW.$