For the circuit of FIGURE EX32.32, Find VR and VL at resonance.
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31. Alternating Current
Series LRC Circuits
Problem 48
Textbook Question
An ac voltage source is connected in series with a 2.0-μF capacitor and a 750-Ω resistor. Using a digital ac voltmeter, the voltage source is measured to be 4.0 V rms, and the voltages across the resistor and across the capacitor are found to be 3.0 V rms and 2.7 V rms, respectively. Determine the frequency of the ac voltage source. Why is the voltage measured across the voltage source not equal to the sum of the voltages measured across the resistor and across the capacitor?

1
Step 1: Understand the problem. The circuit consists of an AC voltage source connected in series with a resistor and a capacitor. The given values are: capacitance \( C = 2.0 \mu F \), resistance \( R = 750 \Omega \), source voltage \( V_{source} = 4.0 \text{ V rms} \), voltage across the resistor \( V_R = 3.0 \text{ V rms} \), and voltage across the capacitor \( V_C = 2.7 \text{ V rms} \). The goal is to find the frequency \( f \) of the AC source and explain why the source voltage is not the arithmetic sum of \( V_R \) and \( V_C \).
Step 2: Recall the relationship between voltages in an AC series circuit. In such a circuit, the voltages across the resistor and capacitor are not simply added algebraically because they are out of phase. The total voltage is the vector sum of \( V_R \) and \( V_C \), calculated using the Pythagorean theorem: \( V_{source} = \sqrt{V_R^2 + V_C^2} \). Verify this relationship using the given values.
Step 3: Calculate the capacitive reactance \( X_C \), which is given by \( X_C = \frac{1}{2 \pi f C} \). Rearrange this formula to solve for the frequency \( f \): \( f = \frac{1}{2 \pi X_C C} \). To find \( X_C \), use Ohm's law for the capacitor: \( V_C = I X_C \), where \( I \) is the current in the circuit. The current can also be found using \( V_R = I R \), so \( I = \frac{V_R}{R} \). Substitute \( I \) into the equation for \( X_C \).
Step 4: Substitute the known values into the equations. First, calculate the current \( I \) using \( I = \frac{V_R}{R} \). Then, use \( X_C = \frac{V_C}{I} \) to find the capacitive reactance. Finally, substitute \( X_C \) and \( C \) into the formula for \( f \) to determine the frequency of the AC source.
Step 5: Explain why the source voltage is not the arithmetic sum of \( V_R \) and \( V_C \). In an AC circuit, the resistor's voltage is in phase with the current, while the capacitor's voltage lags the current by 90°. This phase difference means the voltages are not aligned and must be combined as vectors, not scalars. The total voltage is the resultant of these two perpendicular components, calculated using the Pythagorean theorem.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
AC Voltage and RMS Values
Alternating current (AC) voltage varies sinusoidally over time, and its effective value is often expressed in root mean square (RMS) terms. The RMS value of an AC voltage is equivalent to a direct current (DC) voltage that would deliver the same power to a load. In this question, the source voltage is given as 4.0 V rms, which indicates the effective voltage supplied by the AC source.
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RMS Current and Voltage
Impedance in AC Circuits
In AC circuits, impedance is the total opposition to current flow, combining resistance and reactance. The impedance of a capacitor varies with frequency, given by the formula Z = 1/(ωC), where ω is the angular frequency and C is the capacitance. The presence of both a resistor and a capacitor in series means that the total voltage across the circuit does not simply equal the sum of the individual voltages due to the phase difference between the current and the voltages across the components.
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Impedance in AC Circuits
Voltage Division in AC Circuits
Voltage division in AC circuits occurs when the total voltage is distributed across components based on their impedances. The voltage across each component is determined by the ratio of its impedance to the total impedance of the circuit. In this case, the measured voltages across the resistor and capacitor do not add up to the source voltage because of the phase difference and the reactive nature of the capacitor, which affects how voltages are distributed in the circuit.
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Impedance in AC Circuits
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