Answer:
13.6mm
Explanation:
We consider diameter to be a chord that runs through the center point of the circle. It is considered as the longest possible chord of any circle. The center of a circle is the midpoint of its diameter. That is, it divides it into two equal parts, each of which is a radius of the circle. The radius is half the diameter.
See attachment for the step by step solution of the problem
Answer:
the graph and the answer can be found in the explanation section
Explanation:
Given:
Network rated voltage = 24 kV
Impedance of network = 0.07 + j0.5 Ω/mi, 8 mi
Rn = 0.07 * 8 = 0.56 Ω
Xn = 0.5 * 8 = 4 Ω
If the alternator terminal voltage is equal to network rated voltage will have
Vt = 24 kV/√3 = 13.85 kV/phase
The alternative current is
The impedance Zn is
The voltage drop is
rac = 1.2rdc = 1.2 * 7.476 = 8.97 Ω
The effective armature resistance is
The induced voltage for leading power factor is
if cosθ = 0.5
if cosθ= 0.6
EF = 12790.8 V
if cosθ = 0.7
EF = 13731.05 V
if cosθ = 0.8
EF = 14741.6 V
if cosθ = 0.9
EF = 15809.02 V
if cosθ = 1
EF = 13975.6 V
The voltage regulation is
For each value:
if cosθ = 0.5
voltage regulation = -13.8%
if cosθ = 0.6
voltage regulation = -7.6%
if cosθ = 0.7
voltage regulation = -0.85%
if cosθ = 0.8
voltage regulation = 6.4%
if cosθ = 0.9
voltage regulation = 14%
if cosθ = 1
voltage regulation = 0.9%
the graph is shown in the attached image
for 10% of regulation the power factor is 0.81
Answer:
Explanation:
Given that:
x(t) = 10 sin(10t) . sin (15t)
the objective is to find the power and the rms value of the following signal square.
Recall that:
sin (A + B) + sin(A - B) = 2 sin A.cos B
x(t) = 10 sin(15t) . cos (10t)
x(t) = 5(2 sin (15t). cos (10t))
x(t) = 5 × ( sin (15t + 10t) + sin (15t-10t)
x(t) = 5sin(25 t) + 5 sin (5t)
From the knowledge of sinusoidial signal Asin (ωt), Power can be expressed as:
For the number of sinosoidial signals;
Power can be expressed as:
As such,
For x(t), Power
For the number of sinosoidial signals;
For x(t), the RMS value is as follows: