ch03_ssm - Simple Resistive Circuits 3 Assessment Problems...

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3 Simple Resistive Circuits Assessment Problems AP 3.1 Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains. Begin by combining the 6Ω resistor and the 10 Ω resistor in series: 6Ω+10Ω = 16Ω Now combine this 16 Ω resistor in parallel with the 64 Ω resistor: 16 Ω k 64 Ω = (16)(64) 16+64 = 1024 80 =12 . 8Ω This equivalent 12 . resistor is in series with the 7 . 2Ω resistor: 12 . 8Ω+7 . 2Ω=20Ω Finally, this equivalent 20 Ω resistor is in parallel with the 30 Ω resistor: 20 Ω k 30 Ω = (20)(30) 20+30 = 600 50 =12Ω Thus, the simpli±ed circuit is as shown: 3–1

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3–2 CHAPTER 3. Simple Resistive Circuits [a] With the simplifed circuit we can use Ohm’s law to fnd the voltage across both the current source and the 12 Ω equivalent resistor: v = (12 Ω)(5 A )=60 V [b] Now that we know the value oF the voltage drop across the current source, we can use the Formula p = vi to fnd the power associated with the source: p = (60 V )(5 A )= 300 W Thus, the source delivers 300 W oF power to the circuit. [c] We now can return to the original circuit, shown in the frst fgure. In this circuit, v =60 V, as calculated in part (a). This is also the voltage drop across the 30 Ω resistor, so we can use Ohm’s law to calculate the current through this resistor: i A = 60 V 30 Ω =2 A Now write a KCL equation at the upper leFt node to fnd the current i B : 5 A + i A + i B =0 so i B =5 A i A A 2 A =3 A Next, write a KVL equation around the outer loop oF the circuit, using Ohm’s law to express the voltage drop across the resistors in terms oF the current through the resistors: v +7 . 2 i B +6 i C +10 i C So 16 i C = v 7 . 2 i B V (7 . 2)(3) = 38 . 4 V Thus i C = 38 . 4 16 . 4 A Now that we have the current through the 10 Ω resistor we can use the Formula p = Ri 2 to fnd the power: p 10 Ω = (10)(2 . 4) 2 =57 . 6 W AP 3.2 [a] We can use voltage division to calculate the voltage v o across the 75 k resistor: v o (no load) = 75 , 000 75 , 000+25 , 000 (200 V ) = 150 V
Problems 3–3 [b] When we have a load resistance of 150 k then the voltage v o is across the parallel combination of the 75 k resistor and the 150 k resistor. First, calculate the equivalent resistance of the parallel combination: 75 k k 150 k Ω= (75 , 000)(150 , 000) 75 , 000 + 150 , 000 =50 , 000 Ω = 50 k Now use voltage division to ±nd v o across this equivalent resistance: v o = 50 , 000 50 , 000+25 , 000 (200 V ) = 133 . 3 V [c] If the load terminals are short-circuited, the 75 k resistor is effectively removed from the circuit, leaving only the voltage source and the 25 k resistor. We can calculate the current in the resistor using Ohm’s law: i = 200 V 25 k =8 mA Now we can use the formula p = Ri 2 to ±nd the power dissipated in the 25 k resistor: p 25 k = (25 , 000)(0 . 008) 2 =1 . 6 W [d] The power dissipated in the 75 k resistor will be maximum at no load since v o is maximum. In part (a) we determined that the no-load voltage is 150 V, so be can use the formula p = v 2 /R to calculate the power: p 75 k (max) = (150) 2 75 , 000 =0 . 3 W AP 3.3 [a] We will write a current division equation for the current throught the 80Ω

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This note was uploaded on 04/20/2011 for the course EECE 222 taught by Professor Amsddasgf during the Spring '11 term at American University in Bulgaria.

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ch03_ssm - Simple Resistive Circuits 3 Assessment Problems...

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