Answer:
Explanation:
plate separation = 2.3 x 10⁻³ m
capacity C₁ = ε A / d
= ε A / 2.3 x 10⁻³
C₂ = ε A / 1.15 x 10⁻³
=
a ) when charge remains constant
energy =
q is charge and C is capacity
energy stored initially E₁=
energy stored finally E₂ =
=
=
=
= 4.19 J
b )
In this case potential diff remains constant
energy of capacitor = 1/2 C V²
energy is proportional to capacity as V is constant .
= 16.76 .
Answer:
(a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
Explanation:
Given that,
Power factor = 0.6
Power = 600 kVA
(a). We need to calculate the reactive power
Using formula of reactive power
...(I)
We need to calculate the
Using formula of
Put the value into the formula
Put the value of Φ in equation (I)
(b). We draw the power triangle
(c). We need to calculate the reactive power of a capacitor to be connected across the load to raise the power factor to 0.95
Using formula of reactive power
We need to calculate the difference between Q and Q'
Put the value into the formula
Hence, (a). The reactive power is 799.99 KVAR.
(c). The reactive power of a capacitor to be connected across the load to raise the power factor to 0.95 is 790.05 KVAR.
