9. A voltmeter and frequency meter connected across the terminals of a single-phase alternator running on no load gave readings of 777,7V and 50Hz, respectively. Determine
9.1 the mathematical equation describing this open-circuit voltage
9.2 the average value of this emf
9.3 the times needed for the emf ro reach 975V during the first half coffee rotation.
9.4 the magnitude if the emf 45° after the beginning of q cycle
9.5the magnitude of the emf 1,25ms after the beginning of a cycle
9.1 To describe the open-circuit voltage, we can use the equation:
V(t) = Vm sin(ωt)
Where V(t) is the instantaneous voltage at time t, Vm is the maximum voltage (in this case, 777.7V), ω is the angular frequency (2πf, where f is the frequency, in this case, 50Hz), and t is the time.
So the mathematical equation describing this open-circuit voltage is:
V(t) = 777.7 sin(2π(50)t)
9.2 The average value of the emf can be found by integrating the voltage equation over a complete cycle and dividing by the period.
The period (T) is the reciprocal of the frequency, so T = 1/50 = 0.02 seconds.
The average value (V_avg) is:
V_avg = (1/T) ∫[0 to T] V(t) dt
V_avg = (1/0.02) ∫[0 to 0.02] 777.7 sin(2π(50)t) dt
V_avg = (1/0.02) [ -777.7/(2π(50)) cos(2π(50)t) ] evaluated from 0 to 0.02
V_avg = (1/0.02) [ -777.7/(2π(50)) cos(2π) - (-777.7/(2π(50)) cos(0) ]
V_avg = (1/0.02) [ -777.7/(2π(50)) - (-777.7/(2π(50)) cos(0) ]
V_avg = 0
So the average value of the emf is 0V.
9.3 To find the time needed for the emf to reach 975V during the first half-cycle, we can solve the voltage equation for t:
975 = 777.7 sin(2π(50)t)
Dividing both sides by 777.7:
sin(2π(50)t) = 975/777.7
Taking the inverse sine (sin⁻¹) of both sides:
2π(50)t = sin⁻¹(975/777.7)
Dividing both sides by 2π(50):
t = sin⁻¹(975/777.7) / (2π(50))
Calculate the expression on the right side to find the value of t.
9.4 To find the magnitude of the emf 45° after the beginning of a cycle, we can use the voltage equation:
V(t) = Vm sin(ωt)
Replace t with the time at 45° (π/4 radians):
V(π/4) = 777.7 sin(2π(50)(π/4))
Calculate the expression on the right side to find the magnitude of the emf at 45°.
9.5 To find the magnitude of the emf 1.25ms after the beginning of a cycle, we can use the voltage equation:
V(t) = Vm sin(ωt)
Replace t with the time at 1.25ms (0.00125 seconds):
V(0.00125) = 777.7 sin(2π(50)(0.00125))
Calculate the expression on the right side to find the magnitude of the emf at 1.25ms.