Question

For a balanced three phase, three wire system with input power of 10kW, at 0.9 power factor, the readings on both wattmeter are ————– respectively​

Answer 1

Given that,

Input power = 10 kW

Power factor $\cos\phi = 0.9$

Suppose, the options are

1) 7KW, 3Kw 2) 6350W, 3650W 3 5000W, 5000W 4) 7600W, 1200W

We need to calculate the value of Φ

Using formula of power factor

$\cos\phi=0.9$

$\phi=\cos^{-1}0.9$

$\phi=25.8^{\circ}$

We need to calculate the readings on both wattmeter

Using formula of wattmeter

$\tan\phi=\sqrt{3}(\dfrac{w_{1}-w_{2}}{w_{1}+w_{2}})$

Put the value into the formula

$\dfrac{w_{1}-w_{2}}{w_{1}+w_{2}}=\dfrac{\tan25.8}{\sqrt{3}}$

$\dfrac{w_{1}-w_{2}}{w_{1}+w_{2}}=0.27$....(I)

For option 1,

$w_{1}= 7\ kW$

$w_{2}= 3\ kW$

Put the value of w₁ and w₂ into the formula (I)

$\dfrac{7-3}{7+3}=0.27$

L.H.S≠R.H.S

For option 2,

$w_{1}= 6350\ kW$

$w_{2}= 3650\ kW$

Put the value of w₁ and w₂ into the formula (I)

$\dfrac{6350-3650}{6350+3650}=0.27$

L.H.S=R.H.S

For option 3,

$w_{1}= 5000\ kW$

$w_{2}= 5000\ kW$

Put the value of w₁ and w₂ into the formula (I)

$\dfrac{5000-5000}{5000+5000}=0.279$

L.H.S≠R.H.S

For option 4,

$w_{1}= 7600\ kW$

$w_{2}= 1200\ kW$

Put the value of w₁ and w₂ into the formula (I)

$\dfrac{7600-1200}{7600+1200}=0.279$

L.H.S≠R.H.S

Hence, option 2 is correct.