Chapter 2_Basic Circuit Analysis

# Chapter 2_Basic Circuit Analysis - The Analysis and Design...

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The Analysis and Design of Linear Circuits Seventh Edition 2 Basic Circuit Analysis 2.1 Exercise Solutions Exercise 2–1. A 6-V lantern battery powers a light bulb that draws 3 mA of current. What is the resistance of the lamp? How much power does the lantern use? Using Ohm’s law, we have v = iR or R = v i , so we can compute the resistance as R = 6 V 3 mA = 2 kΩ. The power is p = vi = (6 V)(3 mA) = 18 mW. Exercise 2–2. What is the maximum current that can flow through a 1 8 -W, 6.8-kΩ resistor? What is the maximum voltage that can be across it? The resistor can dissipate up to 0.125 W of power. We have p MAX = i 2 MAX R , which we can solved for i MAX and then substitute in values for the power and resistance i MAX = radicalbigg p MAX R = radicalbigg 0 . 125 6800 = 4 . 2875 mA Similarly, we can use p MAX = v 2 MAX R to solve for the maximum voltage as follows: v MAX = radicalbig R p MAX = radicalbig (6800)(0 . 125) = 29 . 155 V Exercise 2–3. A digital clock is a voltage that switches between two values at a constant rate that is used to time digital circuits. A particular clock switches between 0 V and 5 V every 10 μ s. Sketch the clock’s i - v characteristics for the times when the clock is at 0 V and at 5 V. When the clock has a value of 0 V, its voltage is constant and zero for a wide range of currents. In this case, the i - v characteristic is a vertical line at 0 V. Likewise, when the clock has a value of 5 V, the voltage is constant at 5 V for a wide range of currents. In this case, the i - v characteristic is a vertical line at 5 V. Exercise 2–4. Refer to Figure 2–12. Figure 2–12 (a). Write KCL equations at nodes A, B, C, and D. KCL states that the sum of the currents entering a node is zero at every instant. As we sum the currents at a node, if the current enters that node, it is positive and if the current leaves the node, it is negative. At node A, both currents i 1 and i 2 are leaving the node, so the equation is i 1 i 2 = 0. At node B, current i 2 enters the node and currents i 3 and i 4 leave the node, so we have i 2 i 3 i 4 = 0. At node C, current i 4 enters the node and currents i 5 and i 6 leave the node, so we have i 4 i 5 i 6 = 0. At node D, currents i 1 , i 3 , i 5 , and i 6 enter the node, so we have i 1 + i 3 + i 5 + i 6 = 0. (b). Given i 1 = 1 mA, i 3 = 0 . 5 mA, i 6 = 0 . 2 mA, find i 2 , i 4 , and i 5 . Applying the KCL equation for node A, we can find i 2 = i 1 = 1 mA. Applying the KCL equation for node B, we have i 4 = i 2 i 3 = 1 0 . 5 = 0 . 5 mA. Finally, applying the KCL equation for node C, we have i 5 = i 4 i 6 = 0 . 5 0 . 2 = 0 . 3 mA. Solution Manual Chapter 2 Page 2-1

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The Analysis and Design of Linear Circuits Seventh Edition Figure 2–14 Exercise 2–5. Find the voltages v x and v y in Figure 2–14. To find v x , write the KVL equation around Loop 1 as v x + 2 + 6 = 0 and solve for v x = +8 V. To find v y , write the KVL equation around Loop 2 as v y + 1 6 = 0 and solve for v y = +5 V.
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• Fall '15
• Resistor, Electrical resistance, Electrical impedance, Thévenin's theorem, Series and parallel circuits

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