The impedance on the secondary side of the lh

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Unformatted text preview: 5(Isc + I/2) – 5(Isc – 5Ω I/2) = 0. Again, the terms involving I cancel out. Hence, Isc 5Ω = 1∠0° A, and ZTh = 10∠0 o = 10 Ω. It follows that for 1∠0 o 10∠0° V + Isc + I/2 ISC – I/2 I I/2 j2 Ω Isc – maximum power transfer, Z = 10 Ω. The power dissipated in the 1:2 2 ⎛V ⎞ 1 load is ⎜ Th ⎟ ⎜ ⎟ 4 × 10 = 1.25 W. ⎝ 2⎠ 7/7 If the independent voltage source is replaced by a short circuit, the impedance on the primary side is (5 + j10) Ω and ZTh = 4(5 + j10) = 20 + j40 Ω. j10 Ω 5Ω 3. Determine the impedance seen by the source, 4∠0° V assuming a = 2. 5Ω + – -j5 Ω Solution: Reflection of the (5 – j5) Ω 1:a 2:1 through the RH transformer gives (20 – j20) Ω. The impedance on the secondary side of the LH transformer is (25 – j10) Ω. Reflected to the primary side, this becomes (25 – j10)/a2 Ω. 4. If vsrc = 10cos(1,000t) V, determine the energy stored in the circuit in the sinusoidal steady state at t = 0, assuming C = 1 μF. 1H Solution: At t = 0, the voltage across C is 10 V and vsrc + C k = 0.5 – the current through the inductors is zero, being 1H proportional to the integral of vsrc. The energy stored is W = 1 Cv 2 = 50C. 2 -j5 Ω 5. Determine Rx given that I = 0 and R = 2 Ω. 8Ω Solution: Since I = 0, the voltage across Rx is 10 V, and the same 30∠0 − 10∠0 current R o o flows + -j10 Ω – R I Rx + – 10∠0° V 6Ω 30∠0° V through R and Rx. It follows that R R 20 R x = 10 , or R x = . R 2 1:2 + 7. Determine the maximum power that can be delivered to RL, assuming R 2∠0° V – 5∠0° V = 0.5 Ω. + – 1:2 RL Solution: The primary voltage of the upper transformer is always 1 V. On 3/5 open circuit, the source current is zero, the primary voltage is 5 – 1 = 4 V, and VTh = 8 V. On short circuit, the primary voltage of the lower transformer is zero, the source current is (5 – 1)/R and the short circuit current is 2/R. This gives, RTh = 4R. The maximum power delivered is (8)2/(4×4R) = 4/R. RLine 8. Given that the load L consumes 1200 W at 0.8 p.f. lagging and the magnitude of the voltage across L is 300 V rms. Determine the power dissipated in the resistance Rline, if Rline = 0...
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