Answer:
[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]
Explanation:
Given that:
A circuit with a lagging 0.7 pf delivers 1500 watts and 2100VA
Here:
the initial power factor i.e cos θ₁ = 0.7 lag
θ₁ = cos⁻¹ (0.7)
θ₁ = 45.573°
Active power P = 1500 watts
Apparent power S = 2100 VA
What amount of vars must be added to bring the pf to 0.85
i.e the required power factor here is cos θ₂ = 0.85 lag
θ₂ = cos⁻¹ (0.85)
θ₂ = 31.788°
However; the initial reactive power [tex]Q_1[/tex] = P×tanθ₁
the initial reactive power [tex]Q_1[/tex] = 1500 × tan(45.573)
the initial reactive power [tex]Q_1[/tex] = 1500 × 1.0202
the initial reactive power [tex]Q_1[/tex] = 1530.3 vars
The amount of vars that must therefore be added to bring the pf to 0.85
can be calculated as:
[tex]Q_{sh} = P( tan \theta_1 - tan \theta_2)[/tex]
[tex]Q_{sh} = 1500( tan \ 45.573 - tan \ 31.788)[/tex]
[tex]Q_{sh} = 1500( 1.0202 - 0.6197)[/tex]
[tex]Q_{sh} = 1500( 0.4005)[/tex]
[tex]\mathtt{Q_{sh} = 600.75 \ vars}[/tex]
