VI R1 R3 RL R2 För the above circuit, given that V = 30 Volts, %3D h = 28 Amps, %3D R1 5 Ohms, R2 = 6 Ohms, R3 = 5 Ohms, Note you do not need to find R in this probl Th Use the RTh Method l- Short Circuit, to find /sc Find /
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- Number 9 Please show your detailed solution. Number 9 Please show you detailed solution. A 50-ohm resistor is in series with a 40-?F capacitor across a constant potential source of 180 volts. Determine frequency that will give current of 3.13 amp. A. 250 Hz C. 160 Hz B. 200 Hz D. 140 HzFACTS: AC-DC converters take the AC power from wall outlets and convert it to unregulated DC. These power supplies include transformers that change the voltage of the AC that comes through wall outlets, rectifiers to save it from AC to DC and a filter that removes noise from the peaks and troths of the AC power waves.The DC-to-AC Converters are used to charge the batteries in the vehicles. These circuits are mainly used for driving low-power AC motors and are used in a solar power system. The DC to AC converters can be used in dc transmission lines for transmitting power to loads. QUESTION: What do you think is the main reason why we have AC power in our outlets instead of DC power? Explain your answer.Two storagebatteries, A and B, are connected in parallel for charging from a d.c. sourcehaving an open-circuit voltage of25V and an internal resistance of 2.5 .The open-circuit voltage of A is 11V and that of B is 11.5 V; the internalresistances are 0.06and 0.05 respectively. Calculate the initial chargingcurrents.
- for The Gircuit shown in figure if (B =50) tind The values of CIe). and of Then express The change in value by apercentage + 15U RB, = 100Ka R82 = 50 KQ RE 3K2 %3DThe voltage regulation can be equal to zero :when .The t and current is not too highO .The p.f. is leading p.f. for any possible current O .The p.f. is lagging p.f. for rated current OFor the given circuit in Figure below, find the open circuit voltage VTH and the short circuit current IN across the terminal. then, find R TH by using the RTH=VTH/IN equation and find RTH with replacing voltage source with short circuit. Let’s assume we connect RL (load resistance) to the circuit between points A and B, what should be the value of RL in order to have max power transfer.
- B.. What is the load current for the circuit shown in the figure? Please choose one: a. 6.0 mA b. 3.0 mA C. 9.0 mA D. 7.5 mAPlease solve this 2 problems in detail Problem 1: Consider a step up converter with a resistive load of value R=10 Ohms. It is supplied by a DC power source of magnitude Vs=220 V. The switching frequency is 4 kHz and the duty cycle k is first set to 0.5 and the inductance used within the circuit has a value of L=20 mH. The switch is supposed to be ideal. Using the boundary conditions, the minimum and maximum load current values are equal to: Select one : a. 41.65A and 44.695A b. 43.32A and 44.695A c. None of these d. 43.32A and 47.85A Problem 2: A step up DC converter is used to control power flow from a DC votage Vs = 100V to a battery voltage E = 200V. The power transferred to the battery is 20 kw. The current ripple of the inductor is negligible. The duty cycle and the effective load resistance as seen from the input are respectively equal to: Select one: a. None of these b. 0.5, 0.5 Ohms c. 0.33, 0.5 Ohms d. 0.5, 0.33 OhmsVirtual Lab: circuits and Kirchoff’s rules Go to: https://phet.colorado.edu/it/simulation/circuit-construction-kit-dc-virtual-labBuild each of the circuits in the figures, with the designed characteristics.For each of the circuits, show the calculations to find the current and the potential difference in eachelement of the circuit.In building the circuit in Figure 1) through the simulator, adjust the small resistance of the battery tozero and put in series a small resistance as required from the design.For instance, for E1 you will need abattery with V = 6V and you will need to put in series a small resistance of 0.5 Ω to simulate the smallresistance in a battery.Figure 1)Figure 2)
- In the circuit given in the figure, Rk = 1.93Kohm, R1 = 2071.88Kohm, R2 = 1678.12Kohm, R3 = 2.84Kohm, R4 = 0.95Kohm, R5 = 26.02ohm, Ry = 95.53Kohm, VCC = 15.00V, VTN = 2.40V, Kn = Since it is 2.30(mA/V^2), C1= 5.80uF, C2= 5.20uF, C3= 7.73uF, calculate the lower cutoff frequency of the circuit in Hz.Q1. following: For the circuit shown in figure (a). Let V1 = 10Vms, frequency 10KHZ. Draw the A. Impedance diagram. B. Voltages phasor diagram. C. Current phasor diagram. D. Measure the voltage across R1,R2, Vin, and the inductor using the ocillscope using PROTEUS software. E. Measure the voltage in R1, R2, and the inductor using DMMby using PROTEUS software. F. From the above measurments verify your result. R2 1K V1 VSINE R1 L1 2K 1000MShow complete and step by step solution. It is basic electrical engineering subj.