ch02_ssm - Circuit Elements 2 Assessment Problems AP 2.1...

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2 Circuit Elements Assessment Problems AP 2.1 [a] To fnd v g write a KVL equation clockwise around the leFt loop, starting below the dependent source: i b 4 + v g =0 so v g = i b 4 To fnd i b write a KCL equation at the upper right node. Sum the currents leaving the node: i b +8 A so i b = 8 A Thus, v g = 8 4 = 2 V [b] To fnd the power associated with the 8 A source, we need to fnd the voltage drop across the source, v i . To do this, write a KVL equation clockwise around the leFt loop, starting below the voltage source: v g + v i so v i = v g = 2 V Using the passive sign convention, p s =(8 A )( v i )=(8 A )( 2 V )= 16 W Thus the current source generated 16 W oF power. 2–1
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2–2 CHAPTER 2. Circuit Elements AP 2.2 [a] Note from the circuit that v x = 25 V. To Fnd α write a KCL equation at the top left node, summing the currents leaving: 15 A + αv x =0 Substituting for v x , 15 A + α ( 25 V )=0 so α (25 V )=15 A Thus α = 15 A 25 V . 6 A/V [b] To Fnd the power associated with the voltage source we need to know the current, i v . To Fnd this current, write a KCL equation at the top left node, summing the currents leaving the node: αv x + i v so i v = αv x =(0 . 6)( 25) = 15 A Using the passive sign convention, p s = ( i v )(25 V )= ( 15 A )(25 V ) = 375 W . Thus the voltage source dissipates 375 W. AP 2.3 [a] A KVL equation gives v g + v R so v R = v g =1 kV
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Problems 2–3 Note from the circuit that the current through the resistor is i g =5 mA. Use Ohm’s law to calculate the value of the resistor: R = v R i g = 1 kV 5 mA = 200 k Using the passive sign convention to calculate the power in the resistor, p R =( v R )( i g )=(1 kV )(5 mA )=5 W The resistor is dissipating5Wofpower . [b] Note from part (a) the v R = v g and i R = i g . The power delivered by the source is thus p source = v g i g so v g = p source i g = ( 3 W ) 75 mA =40 V Since we now have the value of both the voltage and the current for the resistor, we can use Ohm’s law to calculate the resistor value: R = v g i g = 40 V 75 mA = 533 . 33 Ω The power absorbed by the resistor must equal the power generated by the source. Thus, p R = p source = ( 3 W )=3 W [c] Again, note the i R = i g . The power dissipated by the resistor can be determined from the resistor’s current: p R = R ( i R ) 2 = R ( i g ) 2 Solving for i g , i 2 g = p r R = 480 mW 300 Ω =0 . 0016 so i g = 0 . 0016 = 0 . 04 A mA Then, since v R = v g v R = Ri R = Ri g = (300 Ω)(40 mA )=12 Vs o v g =12 V AP 2.4
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2–4 CHAPTER 2. Circuit Elements [a] Note from the circuit that the current throught the conductance G is i g , flowing from top to bottom (from KCL), and the voltage drop across the current source is v g , positive at the top (from KVL). From a version of Ohm’s law, v g = i g G = 0 . 5 A 50 mS =10 V Now that we know the voltage drop across the current source, we can ±nd the power delivered by this source: p source = v g i g = (10)(0 . 5) = 5 W Thus the current source delivers 5 W to the circuit.
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ch02_ssm - Circuit Elements 2 Assessment Problems AP 2.1...

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