3 6 derive the time domain expression for vc given

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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.

Ask a homework question - tutors are online