Unformatted text preview: 5
I = jI; the reactive power absorbed
j5 + 5 − j5 by the capacitor is 5jI2 = 5I2 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 I2 = 10I2 VAR. The total reactive power absorbed is 10I2 – 5I2 = 5I2 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 timedomain expression for vC,
given that vSRC = 10sin(2,000t) V. 2 mH 2 mH Solution: ωL = 2×103×2×103 = 4 Ω;
1
1
=
= 5 Ω; VSRC = 10∠0°.
3
ωC 2 × 10 × 100 × 10 −6 ix
vSRC +
_ The nodevoltage 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 righthand 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 103×30×103 = 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.
 Spring '07
 RiadChedid

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