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The Code places a maximum open-circuit voltage on lighting equipment in or on homes. This maximum voltage is (600) (750) (1000). (Circle the correct answer.) Where in the NEC is this voltage maximum referenced? _______________

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EBK ELECTRICAL WIRING RESIDENTIAL
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- Question 3 AC Motor Drives [15]Calculate the instantaneous currents delivered by the inverter if the direct axiscurrent required at a particular instant is 8.66A and the quadrature current is5A. Derive all equations for the three currents.arrow_forwardA certain signal f(t) has the following PSD (assume 12 load): Sp (w) = new + 8(w) - 1.5) + (w + 1.5)] (a) What is the mean power in the bandwidth w≤2 rad/see? (b) What is the mean power in the bandwidth -1.9 to 0.99 rad/sec? Paress(w) dw 2ㅈ -arrow_forward(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_forward
- Draw 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_forwardcan you please show working and steps. The answer is 8kohms.arrow_forward
- PSD 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_forwardCan 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_forward
- Can 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_forwardA. 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_forward
- EBK ELECTRICAL WIRING RESIDENTIALElectrical EngineeringISBN:9781337516549Author:SimmonsPublisher:CENGAGE LEARNING - CONSIGNMENT
