Problems_after_Quiz_3

# 3 6 derive the time domain expression for vc given

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Unformatted text preview: 5 I = jI; the reactive power absorbed j5 + 5 − j5 by the capacitor is -5|jI|2 = -5|I|2 VAR. The current in the inductive branch is – j)I = 5 − j5 I = (1 j5 + 5 − j5 2 ∠-45°I; the reactive power absorbed by the inductor is 5| 2 I|2 = 10|I|2 VAR. The total reactive power absorbed is 10|I|2 – 5|I|2 = 5|I|2 VAR. 5. In the circuit shown, L1 consumes I 160 W at 0.8 p.f. lagging and L2 consumes 320 VAR at 0.6 p.f. lagging. Determine I when XC is VSRC chosen for unity power factor, assuming VSRC = 200∠0° V rms. 1/5 + – L1 L2 XC Solution: Ay unity p.f. the total reactance seen by the source is zero and the source applies At only real power. The real power consumed by L2 is 320 × 0.6 = 240 W. The total real 0.8 400 power supplied by the load is 160 + 240 = 400 W. The current is VSRC ∠0 400 ∠0 o A VSRC = o rms. 3Ω 6. Derive the time-domain expression for vC, given that vSRC = 10sin(2,000t) V. 2 mH 2 mH Solution: ωL = 2×103×2×10-3 = 4 Ω; 1 1 = = 5 Ω; VSRC = 10∠0°. 3 ωC 2 × 10 × 100 × 10 −6 ix vSRC + _ The node-voltage method can be applied, + vC _ the circuit being as shown. At the middle node: 5ix 100 μF 3Ω VC/-j5 + (VC – Vx)/j4 + (VC – 10)/j4 = 0 At the right-hand node: j4 Ω (Vx – VC)/j4 + (Vx – 10)/3 = 5Ix = 5(10 – VC)/j4 Solving, VC = 11.98 + j1.44 = 12.1∠6.86°, so that 10∠0° vC = 12.1sin(2,000t + 6.86°) V. 5Ω 7. Derive VTh and ZTh as Vx Ix + _ -j5 Ω vSRC 20 mH 30 mH + 20 mH 50 μF – 10cos(1,000t + 45°) V. Solution: ωL1 = 5Ω 103×30×10-3 = 30 Ω; ωL2 j10 Ω 1:2 c -3 1 1 Ω; = 3 ωC 10 × 50 × 10 −6 10∠45° + j20 Ω -j20 Ω – = 20 Ω; VSRC = 10∠45°. b a = ωM =10 ×20×10 = 20 3 5Ix a seen between terminals ab, given that vSRC = j4 Ω VC d 1:2 b The circuit in the frequency domain will be as shown, where ω(L1 – M) = 10 Ω; ω(L2 – M) = 0 Ω and is omitted. The j20 Ω in parallel with -j20 Ω is effectively an open circuit. The current in the (5 + j10) Ω impedance is zero, Vcd = 10∠45°, and Vab = VTh = 20∠45°. 2/5 Problem 14 I 5Ω 100∠20 Vrms + – + V _ L1 L2 It is given that the complex power of L1 is 5+j10 VA. It is also given that L2 absorbs 20W at lagging power factor of 0.8. What is the...
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## This test prep was uploaded on 03/11/2014 for the course EECE 210 taught by Professor Riadchedid during the Spring '07 term at American University of Beirut.

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