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Reducing complicated RLC circuits 297
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16-9 At A, a complicated parallel RLC circuit; at B, the same circuit simplified.
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16-10 A series-parallel RLC nightmare.
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298 RLC circuit analysis Another way to find the complex impedance here would be to actually build the circuit, connect a signal generator to it, and measure R and X directly with an impedance bridge. Because the proof of the pudding is in the eating, a performance test must eventually be done anyway, no matter how sophisticated the design theory. Engineers have to build things that work!
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Ohm s law for ac circuits
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Ohm s Law for a dc circuit is a simple relationship among three variables: current (I), voltage (E), and resistance (R). The formulas, again, are I E R E/R IR E/I
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In ac circuits containing negligible or zero reactance, these same formulas apply, as long as you are sure that you use the effective current and voltage.
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The effective value for an ac sine wave is the root-mean-square, or rms, value. You learned about this in chapter 9. The rms current or voltage is 0.707 times the peak amplitude. Conversely, the peak value is 1.414 times the rms value. If you re told that an ac voltage is 35 V, or that an ac current is 570 mA, it is generally understood that this refers to a sine-wave rms level, unless otherwise specified.
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Purely resistive impedances
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When the impedance in an ac circuit is such that the reactance X has a negligible effect, and that practically all of the current and voltage exists through and across a resistance R, Ohm s Law for an ac circuit is expressed as I E Z E/Z IZ E/I
where Z is essentially equal to R, and the values I and E are rms current and voltage.
Complex impedances
When determining the relationship among current, voltage and resistance in an ac circuit with resistance and reactance that are both significant, things get interesting. Recall the formula for absolute-value impedance in a series RLC circuit, Z2 R2 X2
so Z is equal to the square root of R2 X2 . This is the length of the vector R jX in the complex impedance plane. You learned this in chapter 15. This formula applies only for series RLC circuits.
Ohm s law for ac circuits 299 The absolute-value impedance for a parallel RLC circuit, in which the resistance is R and the reactance is X, is defined by the formula: Z2 (RX)2/(R2 X2) X2.
Thus, Z is equal to RX divided by the square root of R2
Problem 16-20
A series RX circuit (Fig. 16-11) has R 50.0 of resistance and X tance, and 100 Vac is applied. What is the current 50.0 of reac-
16-11 A series RX circuit. Notation is discussed in the text.
First, calculate Z2 R2 X2 50.02 ( 50.0)2 2500 2500 square root of 5000, or 70.7 . Then I E/Z 100/70.7 1.41 A.
5000; Z is the
Problem 16-21
What are the voltage drops across the resistance and the reactance, respectively, in the above problem The Ohm s Law formulas for dc will work here. Because the current is I 1.41 A, the voltage drop across the resistance is equal to ER IR 1.41 50.0 70.5 V. The voltage drop across the reactance is the product of the current and the reactance: EX IX 1.41 ( 50.0) 70.5. This is an ac voltage of equal magnitude to that across the resistance. But the phase is different. The voltages across the resistance and the reactance a capacitive reactance in this case, because it s negative don t add up to 100. The meaning of the minus sign for the voltage across the capacitor is unclear, but there is no way, whether you consider this sign or not, that the voltages across the resistor and capacitor will arithmetically add up to 100. Shouldn t they In a dc circuit, yes; in an ac circuit, generally, no. In a resistance-reactance ac circuit, there is always a difference in phase between the voltage across the resistive part and the voltage across the reactive part. They always add up to the applied voltage vectorially, but not always arithmetically. You don t need to
300 RLC circuit analysis be concerned with the geometry of the vectors in this situation. It s enough to understand that the vectors don t fall along a single line, and this is why the voltages don t add arithmetically.
Problem 16-22
A series RX circuit (Fig. 16-11) has R 10.0 and a net reactance X 40.0 . The applied voltage is 100. What is the current Calculate Z2 R2 X2 100 1600 1700; thus, Z 41.2. Therefore, I E/Z 100/41.2 2.43 A. Note that the reactance in this circuit is inductive, because it is positive.
Problem 16-23
Problem 16-24
A parallel RX circuit (Fig. 16-12) has resistance R 30 ohms and a net reactance X 20 . The supply voltage is 50 V. What is the total current drawn from the supply Find the absolute-value impedance, remembering the formula for parallel circuits: Z2 (RX)2 /(R2 X2) 360,000/1300 277. The impedance Z is the square root of 277, or 16.6 . The total current is therefore I E/Z 50/16.6 3.01 A.
Problem 16-25
What is the current through R above Through X The Ohm s Law formulas for dc will work here. For the resistance, IR 2.5 A. 1.67 A. For the reactance, IX E/X 50/( 20) E/R 50/30
These currents don t add up to 3.01 A, the total current, whether the minus sign is taken into account, or not. It s not really clear what the minus sign means, anyhow. The reason that the constituent currents, IR and IX, don t add up to the total current, I, is the same as the reason the voltages don t add up in a series RX circuit. These currents are actually 2D vectors; you re seeing them through 1D glasses.
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