Compute for the following and draw the power triangle a. Real power b. Reactive power c. Apparent power II 10nF 1₂ E (~ IKR 474F TOOMM 4.1k0 F2 E = 25 sin(350t +20) Is -C₂ F3
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- The real power delivered by a source to two impedances, Z1=4+j5 and Z2=10 connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current.Two sinusoidal voltages of the same frequency have rms values of 8 V and 3 V. What is the smallest rms value that the sum of these voltages could have? The largest? Justify your answers.Two alternating voltages are given by v1 = 15sinωt volts and v2 = 25 sin (ωt – π/6) volts. Plot both functions on the same axes and hence determine a sinusoidal expression for the resultant vr = v1 + v2. Check your answer using an analytical method. Your manager has asked you to analyse the variation in in results between the two methods
- in the figure below, find a) The p.d across j5 ohmn inductance b)The p.d across -j10 ohmn capacitance c) the current in 20 ohmn resitance d) the current in (12+j160 ohmn impedance e) the complex power for (12+j16) ohmn impedanceAn industrial installation has a power factor of 0.85 in backward. The demand is 150 KVA, and the operating voltage is 4800 V at 60 Hz. What should be done to bring the power factor to 0.95 in lagging? Indicate what you would connect and with what three-phase power. If the connection of the element to be added is in delta, indicate its reactance and the value of its capacitance or inductance. capacitance or inductance.If a current 5.0 A lags the applied voltage 230 V by an angle ϕ, it can be resolved into two components, OA in phase with the voltage and OB lagging by 90°. If the phasor diagram in to a circuit possessing resistance and inductance in series, OA and OB must not be labelled and respectively. It has become the practice to say that the power factor is laggingwhen the current lags the supply voltage, and leading when the current leads the supply voltage. This means that the supply voltage is regarded as the reference quantity. (a) A 30 μ F capacitor is connected across the applied voltage, 50 Hz supply. Calculate the reactance of 30 μ F capacitor connected across applied voltage (b) the reactance of the capacitor; (c) the current.
- Read A 70-Vac source has the following waveform. Determine:a. Vpkb. Vpk-pkc. Vrmsd. Periode. Frequencyf. Angular Velocityg. Phase Angleh. equation of the waveform (in time domain)i. the instantaneous voltage when t = 120 msj. the angle (1st occurrence) after t = 0 when the voltage is +80 Vk. the time (2nd occurrence) after t = 0 when the voltage is –10 VAnswers:a. 98.9949 Vb. 197.9899 Vc. 70 Vd. 100 mse. 10 Hzf. 20π rad/secg. 125.0876°h. v(t) = 98.9949sin(20πt + 125.0876°)i. -29.0880 Vj. 0.99959°k. 63.643 ms Give the solution for J and K ONLY ans (0.99959 degrees and 63.643 ms respectively)A 60-Hz 220-V-rms source supplies power to a load consisting of a resistance in series with an inductance. The real power is 1500 W, and the apparent power is 2500 VAR. Determine the value of the resistance and the value of the inductance.A 50 Hz, alternating voltage of 150 V (rms) is applied independently to (a) Resistance of 10 Ω, (b) Inductance of 0.2 H, (c) Capacitance of 50 μF. Find the expression for the instantaneous current in each case. Draw the phasor diagram in each case.
- An industrial process requires a power of 10 kW with a lagging power factor of 0.6 in a single-phase system. It is desired to add a parallel capacitance to raise the power factor to unity. For a line voltage of 240 V rms, determine the value of capacitance required.A current of 4 A flows when a neon lightadvertisement is supplied by a 110-V rms powersystem. The current lags the voltage by 60◦. Find thepower dissipated by the circuit and the power factor.A 60-Hz 240-V-rms source supplies power to a load consisting of a resistance in series with an inductance. The real power is 1500W, and the apparent power is 2500 VA. Find the impedance and power factor in the system.