
EBK ELECTRICAL ENGINEERING
7th Edition
ISBN: 8220106714201
Author: HAMBLEY
Publisher: YUZU
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What is the current flowing through the load resistor, RL (in ARMS)? How much power does the voltage source, V1, provide to the circuit? The magnitude of V1 is given in VRMS.
We wish to power an extremely bright light to communicate with a neighbor using morse code. We let the system run 24/7, but we swap out the battery every 24 hours for a fully charged one and recharge the drained battery with a solar charger. Based on the signal we are sending, the light draws 2.5 A of current for 2 seconds every 5 seconds. As well, the computer sending the signal to the light continuously draws 120 mA. A 12 V lead acid battery is used to provide the power. To preserve the longevity of the battery we wish to keep the lower limit of the SoC to 75%. (a) What is the minimum battery capacity in Ah required? (b) If a 60 W 12 V solar panel was used to recharge the battery, noting that we will keep the lower SoC to 75%, how many hours of adequate sunlight would be needed each day? (c) If the solar charger malfunctions, and we are forced to use one battery without recharging, what would the battery’s SoC be after 2 days?
1. In the following unity feedback system, we have
G(s)
=
R(s) +
K(s + 1)
s(s + 2)(s +5)
G(s)
C(s)
use Routh-Hurwitz stability criterion to find the range of K for the stability of the system.
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- 4. Discussion: Compare between theoretical effect of KD at first order and second order systems regarding steady-state errors and transient responses with the practical obtained results whenever applying step input signal. In Experiment Derivative Controller 55-82arrow_forwardFor the state space model, find the following: 1. Identify the state-space matrices A,B,C, and D. 2. Compute the transfer function G(s) analytically (by hand). 3. Solve for x(t) given a step input u(t) =1 and zero initial conditions (use transition matrix). 4. Use MATLAB to compute the transfer function. 5. Plot the step response using MATLAB. [X] = 71+0u y = x1 + x2 -2 นarrow_forwardA DPSK has the following data input: d(n) =101011010001 1. Find the output coded sequence and the carrier phase. 2. Recover the input data from the output coded sequence.arrow_forward
- Q9 A single-phase transformer, 2500 / 250 V, 50 kVA, 50 Hz has the following parameters, the Primary and secondary resistances are 0.8 ohm and 0.012 ohm respectively, the primary and secondary reactance are 4 ohm and 0.04 ohm respectively and the transformer gives 96% maximum efficiency at 75% full-load. The magnetizing component of-load current is 1.2 A on 2500 V side. 1- Draw the equivalent circuit referred to primary (H.V side) and inserts all the values in it 2- Find out Ammeter, voltmeter and wattmeter readings on open-circuit and short-circuit test. If supply is given to 2500 V side in both cases. Ans. O.C. Test (Vo= 2500 V, lo=1.24 A, Wo=781.25 w) S.C. Test (Vsc =164.924 V, Isc =20 A, Wsc =800 w )arrow_forwardA modulating signal f(t) is bandlimited to 5.5 kHz is sampled at a rate of 15000 samples/sec. The samples are quantized into 1024 levels. Calculate transmission bandwidth if the following modulation types are used for signal transmission: 1-ASK, QAM 2-QPSK, 8-PSK 3-FSK, 8-FSK with Af = 20 kHzarrow_forwardQ10 The full-load copper loss on the H.V. side of 100KVA, 11000/317 V, single-phase transformer is 0.62 kw and on the L.V. side is 0.48 kW. i) Calculate R1, R2 in ohms ii) Find X1,X2,if the percentage equivalent reactance is 4%, and reactance is divided in same proportion as resistance. Ans, 27.30, 0.175), 0.00482 . (7.5)arrow_forward
- Find the binary sequence, for the following Differential Manchester code.arrow_forwardQ2- What are the parameters and loss that can be determined during open-circuit test of singlephase transformer. Draw the circuit diagram of open-circuit test and explain how can you calculate the Parameters and loss.arrow_forwardQ6- the open circuit and short circuit tests on a 10 KVA, 125/250 v, 50 Hz single phase transformer gave the following results: O.C. Test: 125 V,0.6 A, 50 W ( on L.V.) S.C. Test: 20 V, 40 A, 177.78 W (on H.V. side) Calculate: i) Copper losses on half load ii) Full load efficiency at 0.8 leading p.f. iii) Half load efficiency at 0.8 leading p.f. iv) Regulation at full load at 0.9 leading p.f. Ans: 44.445 W, 97.23%, 97.69%, -1.8015%arrow_forward
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