# 2 since r and g are positive quantities the power

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Unformatted text preview: tor is a nonlinear function of either current or voltage. 2. Since R and G are positive quantities, the power dissipated in a resistor is always positive. Thus, a resistor always absorbs power from the circuit. This conﬁrms the idea that a resistor is a passive element, incapable of generating energy. EXAMPLE 2.1 | v v An electric iron draws 2 A at 120 V. Find its resistance. | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents 32 PART 1 DC Circuits Solution: From Ohm’s law, R= 120 v = = 60 i 2 PRACTICE PROBLEM 2.1 The essential component of a toaster is an electrical element (a resistor) that converts electrical energy to heat energy. How much current is drawn by a toaster with resistance 12 at 110 V? Answer: 9.167 A. EXAMPLE 2.2 In the circuit shown in Fig. 2.8, calculate the current i , the conductance G, and the power p. i + v − 30 V + − 5 kΩ Figure 2.8 For Example 2.2. Solution: The voltage across the resistor is the same as the source voltage (30 V) because the resistor and the voltage source are connected to the same pair of terminals. Hence, the current is i= 30 v = = 6 mA R 5 × 103 The conductance is G= 1 1 = = 0.2 mS R 5 × 103 We can calculate the power in various ways using either Eqs. (1.7), (2.10), or (2.11). p = vi = 30(6 × 10−3 ) = 180 mW or p = i 2 R = (6 × 10−3 )2 5 × 103 = 180 mW or p = v 2 G = (30)2 0.2 × 10−3 = 180 mW PRACTICE PROBLEM 2.2 i 10 kΩ 2 mA | v v Figure 2.9 + v − For the circuit shown in Fig. 2.9, calculate the voltage v , the conductance G, and the power p . Answer: 20 V, 100 µS, 40 mW. For Practice Prob. 2.2 | e-Text Main Menu | Textbook Table of Contents | Problem Solving Workbook Contents CHAPTER 2 Basic Laws 33 EXAMPLE 2.3 A voltage source of 20 sin π t V is connected across a 5-k resistor. Find the current through the resistor and the power dissipated. Solution: i= 20 sin π t v = = 4 sin π t mA R 5 × 103 Hence, p = vi = 80 sin2 π t mW PRACTICE PROBLEM 2.3 A resistor absorbs an instantaneous power o...
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## This note was uploaded on 07/16/2012 for the course KA KA 2000 taught by Professor Bkav during the Spring '12 term at Cambridge.

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