- Ontrol Ac Power As Opposed To A Variable Resistor 17 Suppose A Power System Were Delivering Ac Power To A Resistive L 1 (103 KiB) Viewed 89 times
- Ontrol Ac Power As Opposed To A Variable Resistor 17 Suppose A Power System Were Delivering Ac Power To A Resistive L 2 (154.27 KiB) Viewed 89 times
ontrol AC power, as opposed to a variable resistor? 17. Suppose a power system were delivering AC power to a resistive load drawing 150 amps: I = 150 A Rwire = 0.12 Load AC voltage source 240 VAC Rwire = 0.1 22 Calculate the load voltage, load power dissipation, the power dissipated by the wire resistance (Rcire), and the overall power efficiency (n b). Eload = Pload = Pines - Now, suppose we were to use a pair of perfectly efficient 10:1 transformers to step the voltage up for transmission, and back down again for use at the load. Re-calculate the load voltage. load power, wasted power, and overall efficiency of this system: Rewire = 0.112 w I= 150 A Load e 2400 VAC AC voltage source 240 VAC Rwire = 0.1 22 Eload Pload = Plincs = This moment in time 18. Shown here is a schematic at the exact moment where the
19. An audio power amplifier with an internal impedance of 16 22 needs to power a set of speakers with a combined total impedance of 412. We know that connecting this speaker array directly to the amplifier's output will not result in optimum power transfer, because of the impedance mismatch. Someone suggests using a transformer to match the two disparate impedances. What turns ratio does this transformer need? a Line 120 V 20. Connect the windings of this step-down transformer IN SERIES in an autotransformer configuration with the load (motor), so that it ”boosts" the voltage going to the load by 12 volts, to 132V (HINT: connect the primary and secondary in SERIES, in the proper polarity order so the voltages add): 000000000 12 V Mtr Load 21. Calculate all currents in this autotransformer circuit, assuming perfect (100%) efficiency. The values in the transformer windings may surprise you! Use the following steps: 120 V lllllll W 85 V 500 12 a) Load = Vload/R = mA b) Power Pload=I*V = W c) Isupply (Supply current from 120V source) = Pload/120V = mA d) By KCL, current through lower half of autotransformer = Isupply - Iload = e) Label the values on the schematic, with arrows showing the current directions. mA 22. For the circuit below, determine: R Supply 18k W 3:2 + esupply Rwad V2 Vload 4k 120 V RMS 60Hz a) Referred resistance seen at the primary, rl = b) Primary voltage v1 (NOT the same as esupply)= c) Secondary voltage v2= d) Secondary current i2= e) Primary current il= f) Secondary power to the load PrLoad= g) Source power generated by the supply esupply Ps=