A circuit element consists of a resistor with value 20Ω and inductor with value 10mH connected in series. A voltage of LaTeX: v(
1 answer:
Answer:
8.97 Watt
Explanation:
Resistance, R = 20 ohm
Inductance, L = 10 mH
V(t) = 20 Cos (1000 t + 45°)
Compare with the standard equation
V(t) = Vo Cos(ωt + Ф)
Ф = 45°
ω = 1000 rad/s
Vo = 20 V
Inductive reactance, XL = ωL = 1000 x 0.01 = 10 ohm
impedance is Z.
Z = 22.36 ohm
Apparent power is given by
P = Vrms x Irms
P = 14.144 x 0.634
P = 8.97 Watt
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14 m/s or 50km/h. See the details in the attached picture.
(a) Let's convert the final speed of the car in m/s:
The kinetic energy of the car at t=19 s is
(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
(c) The instantaneous power is given by
where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
The kinetic energy of the car at t=19 s is
(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
(c) The instantaneous power is given by
where F is the force exerted by the engine, equal to F=ma.
So we need to find the acceleration first:
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
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.
