L9 solutions to problems
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The document presents two solutions for calculating power loss in a circuit, with a focus on determining values for ig and rs. The first solution finds a peak power loss of 5 W and average power dissipation of 1 W, while the second solution, using ig = 0.5 A and vg = 8 V, results in a power dissipation of 4 W. Additionally, calculations for average power dissipation are derived using a duty cycle formula.
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L9 solutions to problems
- 1. Solution 1 The basic circuit equation is Subsituting the given values Peak Power loss = Vg Ig = 5 Solving above quadratic eqn for Ig and taking the +ve value we get Ig = 0.7 A. Using this value in (1) Rs = 7 Ω s s g gE R I V 12 1.5 8 8 1.5 ...(1)s g g s gR I I R I 2 5 1.5 8 8 1.5g g g gI I I I 12-Oct-12 1EE-321N, Lec-10
- 2. Solution 1 Contd... Average Power Dissipation Pgav = δ·Pgm = 0.2 × 5 = 1 W 12-Oct-12 2EE-321N, Lec-10
- 3. Solution 2 Given Ig = 500 mA = 0.5 A, Vg/Ig = 16 V/A So, Vg = 16 × 0.5 = 8 V Again using the basic circuit equation Rs is obtained as s s g gE R I V 15 8 14 0.5 s g s g E V R I 12-Oct-12 3EE-321N, Lec-10
- 4. Solution 2 ... contd Power dissipation Pg = Vg Ig = 8 × 0.5 = 4 W Also Pgav = δ·Pgm = Pgm = f ·TON ·Pgm Thus, ONT T 6 0.3 18.75 kHz 19 kHz 4 10 4 f 12-Oct-12 4EE-321N, Lec-10