Concept explainers
Describe how pressure switches are connected to start and stop (a) small motors and (b) large motors.
(a)

Explain how does the pressure switches are connected to start and stop small motors.
Answer to Problem 1SQ
The pressure switches are connected directly to start and stop small motors.
Explanation of Solution
Discussion:
In industrial applications, the pressure switches are used to sense pressure. A large variety of pressure switches are available to control the requirements of hydraulic machines or pneumatic machines.
Single-pole pressure switches are most commonly used and Two-pole pressure switches are used only for few applications.
In small motors, the pressure switches are directly connected to the load circuits when the capacities of the contact are sufficient for the load current.
Conclusion:
Thus, the pressure switches are connected directly to start and stop small motors.
(b)

Explain how does the pressure switches are connected to start and stop large motors.
Answer to Problem 1SQ
The pressure switches are connected series with motor starter and line to start and stop large motors.
Explanation of Solution
Discussion:
In industrial applications, the pressure switches are used to sense pressure. A large variety of pressure switches are available to control the requirements of hydraulic machines or pneumatic machines.
Single-pole pressure switches are most commonly used and Two-pole pressure switches are used only for few applications.
In large motors, the pressure switches alone cannot handle the load circuits safely where it must be connected in series with the motor starter coil and line such that the starter can start and stop large motors safely.
Conclusion:
Thus, the pressure switches are connected in series with motor starter and a line are to start and stop large motors.
Want to see more full solutions like this?
Chapter 10 Solutions
EBK ELECTRIC MOTOR CONTROL
- (75 Marks) JA signal (t) is bond 7)(t)(t) and f(t), are band-limited to 1.2 kHz each. These signals are to be limited to 9.6 kHz, and three other signals transmitted by means of time-division multiplexing. Set up scheme for accomplishing this multiplexing requirement, with each signal sampled at its Nyquist rate. What must be the speed of the commutator (the output but ram-k bit/sec)? the minimum band width? (25 Marks)arrow_forwardDraw the digital modulation outputs, ASK Amplitude Shift Keying) FSK (Frequency Shift Keying) and PSK (Phase Shift Keying). For baseband and carriet frequency as shown 101 wwwwwwwwwwww 010 BASESAND basband CARRIER Carralarrow_forwardplease show full working. I've included the solutionarrow_forward
- can you please show working and steps. The answer is 8kohms.arrow_forwardPSD A certain signal f(t) has the following PSD (assume 12 load): | Sƒ(w) = π[e¯\w\ + 8(w − 2) + +8(w + 2)] (a) What is the mean power in the bandwidth w≤ 1 rad/sec? (b) What is the mean power in the bandwidth 0.99 to 1.01 rad/sec? (c) What is the mean power in the bandwidth 1.99 to 2.01 rad/sec? (d) What is the total mean power in (t)? Pav= + 2T SfLw) dw - SALW)arrow_forwardAn AM modulation waveform signal:- p(t)=(8+4 cos 1000πt + 4 cos 2000πt) cos 10000nt (a) Sketch the amplitude spectrum of p(t). (b) Find total power, sideband power and power efficiency. (c) Find the average power containing of each sideband.arrow_forward
- Can you rewrite the solution because it is unclear? AM (+) = 8(1+ 0.5 cos 1000kt +0.5 ros 2000ks) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. -Jet jooort J11000 t = 4 e jqooort jgoort +4e + e +e j 12000rt. 12000 kt + e +e jooxt igoo t te (w) = 8ES(W- 100007) + 8IS (W-10000) USBarrow_forwardCan you rewrite the solution because it is unclear? AM (+) = 8(1+0.5 cos 1000kt +0.5 ros 2000 thts) = cos 10000 πt. 8 cos wat + 4 cos wit + 4 cos Wat coswet. J4000 t j11000rt $14+) = 45 jqooort +4e + e + e j 12000rt. 12000 kt + e +e +e Le jsoort -; goon t te +e Dcw> = 885(W- 100007) + 8 IS (W-10000) - USBarrow_forwardCan you rewrite the solution because it is unclear? Q2 AM ①(+) = 8 (1+0.5 cos 1000πt +0.5 ros 2000kt) $4+) = 45 = *cos 10000 πt. 8 cos wat + 4 cosat + 4 cos Wat coswet. j1000016 +4e -j10000πt j11000Rt j gooort -j 9000 πt + e +e j sooort te +e J11000 t + e te j 12000rt. -J12000 kt + с = 8th S(W- 100007) + 8 IS (W-10000) <&(w) = USB -5-5 -4-5-4 b) Pc 2² = 64 PSB = 42 + 4 2 Pt Pc+ PSB = y = Pe c) Puss = PLSB = = 32 4² = 8 w 32+ 8 = × 100% = 140 (1)³×2×2 31 = 20% x 2 = 3w 302 USB 4.5 5 5.6 6 ms Ac = 4 mi = 0.5 mz Ac = 4 ५ M2 = =0.5arrow_forward
- A. Draw the waveform for the following binary sequence using Bipolar RZ, Bipolar NRZ, and Manchester code. Data sequence= (00110100) B. In a binary PCM system, the output signal-to-quantization ratio is to be hold to a minimum of 50 dB. If the message is a single tone with fm-5 kHz. Determine: 1) The number of required levels, and the corresponding output signal-to-quantizing noise ratio. 2) Minimum required system bandwidth.arrow_forwardFind Io using Mesh analysisarrow_forwardFM station of 100 MHz carrier frequency modulated by a 20 kHz sinusoid with an amplitude of 10 volt, so that the peak frequency deviation is 25 kHz determine: 1) The BW of the FM signal. 2) The approximated BW if the modulating signal amplitude is increased to 50 volt. 3) The approximated BW if the modulating signal frequency is increased by 70%. 4) The amplitude of the modulating signal if the BW is 65 kHz.arrow_forward
