A motor with a coil that has a series resistance of 100 ohm and an inductance of 30H is powered by a 120 V DC generator. To prevent damage to the motor, a 300 ohm field discharge resistor is connected in parallel with it, as seen in Figure 1. The switch opens at t = 0 s after being in the closed state for a considerable amount of time. Find the voltage at 120 ms through the motor resistor (i.e 100 ohm )..

A motor with a coil that has a series resistance of 100 ohm and an inductance of 30H is powered by a 120 V DC generator. To prevent damage to the motor, a 300 ohm field discharge resistor is connected in parallel with it, as seen in Figure 1. The switch opens at t = 0 s after being in the closed state for a considerable amount of time. Find the voltage at 120 ms through the motor resistor (i.e 100 ohm )..

Electricity for Refrigeration, Heating, and Air Conditioning (MindTap Course List)

10th Edition

ISBN:9781337399128

Author:Russell E. Smith

Publisher:Russell E. Smith

Chapter7: Alternating Current, Power Distribution, And Voltage Systems

Section: Chapter Questions

Problem 7RQ

See similar textbooks

Similar questions

for the circuit shown below If v_in=5 sin2n10t then corner frequency will be Hz 1 UF 2 k Ohm 1 k Ohim 050 060 070 080 Min
Determine the DC currents (B. Ic and le) and de juncion voltages (VBE, VCE and Vca) mA. Ic ma, E mA VBE= V. VCE Vand VCB Use 2 decimal places. RB Use the following values: VBB = 3V RB = 4 kQ Rc = 166 0 Vcc = 19 V BDC = 70 %3D %3D VCC RC VBB
A non-sinusoidal current source (iT=2+1.5sin1000t+ 1cos3000t-0.5sin4000t) A is ?applied to the circuit shown. Find the total power 4Ω iL ww ic iT -J92 J42 55 W O 28 W O 20 W O 39 W O
A non-sinusoidal current source (iT=2+1.5sin10o0t+ 1cos3000t-0.5sin4000t) A is applied to the circuit shown. Find the total power? * İL iT ic -J9Ω J4Q O 55 W 20 W 39 W 28 W
A non-sinusoidal current source (iT=2+1.5sin1000t+ 1cos3000t-0.5sin4000t) A is applied to the circuit shown. Find the total power? * 4Ω iL www ic iT -J92 J42 39 W O 28 W O 55 W O 20 W
Referring to the Figure 1 below, set the value of voltage Vs as: - Vs (rms) = (120+the last 2 digit ID Number) ≤45° V Example: ID No: RE20630; So, s (rms) Vs (rms) = 150/45° V Then, calculate the following parameters: - i) The value of power factor. (State either lagging or leading power factor) ii) The average power supplied by the generator, PT iii) The total reactive power, Qr (State either capacitive or inductive load) iv) The total apparent power, ST v) The complex power, ST (Write your answer in 3 decimal places.) = (120+30)/45° V
QUESTION 3 Add the complex numbers -1.22+j2.44 and -4.25+j5.25 -5.47+j7.69 O 1.08-j6.2 -3.66+j10.5 O 2.04+j5.55
For the Photoflash unit. Vse 9V. RISKO C-1500 uf. R2= 120. The average power dissipated by the lamp is High voltage de supply 6.75W 8.3W 0.675W 10W 0.6W The mutual inductance sign for the following circuit is 352 www Negative R₂ Positive 2 mH 3 mH ell 5 mH www 10 Ω
The voltmeters in the figure measure RMS voltages. If V1 = 13 V and V2 = 8.5V what is V3? C R V V2 V3 O 15.53 V O 4.5 V O 21.5 V O 8.5 V О 13V
A solenoid with an inductance of 0.25 H and a resistance of 100 Q is connected to a 120-V 60-Hz source. The current is: Please check the closest answer that applies. X L 120 V 60 Hz O0.226 2-16.3° A O0.224 2-31.2° A LO
assume that the voltage drop across the resistor ER is 78 v the voltage drop across the capacitor EC is 104 v and the circuit has a total impedance z of 20 ohms the frequency of the ac voltage is 60 hz find the missing values. fine the missing vaules
Sind RT R2 220 R冬12k~ R4 b.