Force: Power distribution in a Center-tapped Transformer

S contains mn, n = 1, 2, 3, 4, 5
S = Earth; m1 = voltage source, V1; m2 = primary coil; m3 = magnetic medium; m4 = Secondary coil; m5 = conducting wire;

Center Tapped Transformer

Figure 8.42 shows a center-tapped Transformer. The following information is given about the transformer:
Primary voltage = 4,800 V;
Secondary voltage of 240 V is split into two 120 V voltages (V2, V3) due to the center tap of the transformer. Voltages and current are rms values.
Three resistive loads (R1, R2, R3) are connected to the transformer.
R1 is connected to the 240 V line
R2 and R3 are connected to each of the 120 V lines.
Determine the power absorbed by each of the loads, if:
power absorbed by R2 = P2
power absorbed by R1 = 5P2
power absorbed by R3 = 1.5P2
current through primary coil I1 = 1.5625 A.

Force: Conservation of Power in an Ideal Transformer

S contains mn, n = 1, 2, 3, 4, 5, 6
S = Earth; m1 = voltage source, v1; m2 = primary coil; m3 = magnetic medium; m4 = Secondary coil; m5 = voltage source, v2; m6 = conducting wire;

Ideal Transformer

Figure 8.35 shows an Ideal Transformer. Show that:
The apparent power of the primary coil equals the apparent power of the secondary coil.

Force: AC Power Factor Correction

S contains mn, n = 1, 2, 3, 4
S = Earth; m1 = voltage source, vs; m2 = Capacitor, L; m3 = Load, L; m4 = Conducting, wire

AC Power Factor Correction

Force: AC Power Triangle

S contains mn, n = 1, 2, 3
S = Earth; m1 = voltage source, v(t); m2 = Load, L; m3 = conducting wire.

AC Power Triangle

Figure 8.14 shows a simple AC circuit. Figure 8.14(a) is the time domain circuit while figure 8.14(b) is its power triangle. Determine:
(a) The power factor, pf
(b) The values of P, Q and S in the power triangle.
v(t) = 16cosωt
i(t) = 4cos(ωt - π/6); ω = 377 rad/sec.

Force: Average Power Of AC Circuits

S contains mn, n = 1, 2, 3
S = Earth; m1 = voltage source, v(t); m2 = Load, L; m3 = conducting wire.

AC Power Average

Figure 8.7 shows a simple AC circuit. Figure 8.7(a) is the time domain circuit while figure 8.7(b) is its phasor form. Determine:
(a) The average power of the circuit in the time domain
(b) The average power of the circuit in the frequency domain
v(t) = Vcos(ωt)
i(t) = Icos(ωt - θ)

Change: Resonance - Sharpness Of Frequency Response

S contains mn, n = 1, 2, 3, 4, 5, 6
S = Earth; m1 = voltage source, Vi; m2 = Capacitor, C; m3 = Inductor, L; m4 = Resistor, R; m5 = voltage source, Vo; m6 = conducting wire.

Band Pass Filter

Figure 7.56 shows a circuit of a simple RCL filter. Determine:
(a) The frequency response of the RCL filter in terms of the natural or resonant frequency.
(b) The bandwidth in terms of the natural or resonant frequency and the quality factor.

Change: RCL Band Pass Filter

S contains mn, n = 1, 2, 3, 4, 5, 6
S = Earth; m1 = voltage source, Vi; m2 = Capacitor, C; m3 = Inductor, L; m4 = Resistor, R; m5 = voltage source, Vo; m6 = conducting wire.

Band Pass Filter

Figure 7.56 shows a circuit of a simple RCL filter. Determine:
(a) The frequency response of the RCL filter in terms of the passband or bandwidth frequencies.
(b) The phasor form of the frequency response in terms of the passband or bandwidth frequencies.

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