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.
Answer:
the correct answer is option C
Answer:
Here the circuit in which a 4Ω resistor resistor is connected in series and two 8Ω resistor resistors are connected in parallel. Also, ammeter and voltmeter connected in series and parallel circuit respectively.
Now,
The maximum power of each resistance is 16 W
The 4Ω resistor is linked in series with the circuit.
so, P o w e r = I
two
R, here i is the current through the resistor resistor R
1 6 = I
two
∗ 4 Ω
i = 2A
Now 2A passes through parallel resistors of 8Ω resistance.
we know that, in parallel, the potential difference must be constant,
the current is divided into two parts, because the same resistance current in each resistance will be half. then the current through each resistor in parallel is
2 A
two
.
= 1 A
So finally the current through the 4Ω resistor = 2 A
current through each 8Ω resistor = 1 A
Explanation:
I hope this answer has helped you
Answer:
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